Chapter 8

Linear spaces

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Section 20Examples of spaces. Basis and dimension[34/91%]
Problem 20.1

Is it possible, by a suitable introduction of the operations of addition and multiplication by a number, to make into a vector space:

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(1)

the empty set

(2)

a set of one element

(3)

a set of two elements?

Problem 20.2

Prove that:

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(1)

if a system of vectors contains a linearly dependent subsystem, then it is linearly dependent

(2)

if a system of vectors is linearly independent, then any of its subsystems is also linearly independent

(3)

if the vectors a1,…,aka_{1}, \ldots , a_{k} are linearly independent, and the vectors a0,a1,…,aka_{0}, a_{1}, \ldots , a_{k} are linearly dependent, then the vector a0a_{0} is a linear combination of the vectors a1,…,aka_{1}, \ldots , a_{k}.

Problem 20.3

Determine whether the given set of vectors in an nn-dimensional space is a vector subspace, and if it is, find its dimension:

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(1)

the set of vectors all of whose coordinates are equal to each other

(2)

the set of vectors whose first coordinate equals 0

(3)

the set of vectors the sum of whose coordinates equals 0

(4)

the set of vectors the sum of whose coordinates equals 1.

Problem 20.4

Determine whether the given set of vectors of geometric space is a vector subspace, and if it is, determine its dimension:

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(1)

the set of vectors of the plane parallel to a given line

(2)

the set of vectors of three-dimensional space perpendicular to a given line

(3)

the set of vectors of the plane whose magnitude does not exceed 1

(4)

the set of vectors of the plane forming angle α\alpha with a given line (0∘⩽α⩽90∘)\left(0^{\circ } \leqslant \alpha \leqslant 90^{\circ }\right).

Problem 20.5

Prove that the set of matrices of size m×nm \times n forms a vector space with respect to the usual operations of addition of matrices and multiplication of a matrix by a number. Find the dimension and a basis of this space Rm×n\mathcal{R}_{m \times n}.

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Problem 20.6

Determine whether the given set of square matrices of order nn is a vector subspace in the space of all square matrices of order nn, and if it is, find its dimension:

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(1)

the set of matrices with zero first row

(2)

the set of diagonal matrices

(3)

the set of upper triangular matrices

(4)

the set of symmetric matrices

(5)

the set of skew-symmetric matrices

(6)

the set of singular matrices.

Problem 20.7

Determine whether the given set of functions on an arbitrary segment [a,b]\left[a, b\right] forms a vector space with respect to the usual operations of addition and multiplication by a number:

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(1)

the set of functions continuous on [a,b]\left[a, b\right]

(2)

the set of functions differentiable on [a,b]\left[a, b\right]

(3)

the set of functions Riemann integrable on [a,b]\left[a, b\right]

(4)

the set of functions bounded on [a,b]\left[a, b\right]

(5)

the set of functions such that sup⁡[a,b]∣f(x)∣⩽1\sup_{\left[a, b\right]} \left|f(x)\right| \leqslant 1

(6)

the set of functions nonnegative on [a,b]\left[a, b\right]

(7)

the set of functions such that f(a)=0f(a) = 0

(8)

the set of functions such that f(a)=1f(a) = 1

(9)

the set of functions such that lim⁡x→a+0f(x)=∞\lim_{x \rightarrow a + 0} f(x) = \infty

(10)

the set of functions monotonically increasing on [a,b]\left[a, b\right]

(11)

the set of functions monotonic on [a,b]\left[a, b\right].

Problem 20.8

Prove that for any natural number nn the given set of functions forms a finite-dimensional vector space; find the dimension and indicate a basis of this space:

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(1)

the set of polynomials of degree at most nn (denoted P(n)\mathcal{P}^{(n)})

(2)

the set of even polynomials of degree at most nn

(3)

the set of odd polynomials of degree at most nn

(4)

the set of trigonometric polynomials of order at most nn, i.e. the set of functions of the form f(t)=a0+a1cos⁡t+b1sin⁡t+…+ancos⁡nt+bnsin⁡ntf(t) = a_{0} + a_{1} \cos t + b_{1} \sin t + \ldots + a_{n} \cos nt + b_{n} \sin nt

(5)

the set of even trigonometric polynomials of order at most nn

(6)

the set of odd trigonometric polynomials of order at most nn

(7)

the set of functions of the form f(t)=eαt(a0+a1cos⁡t++b1sin⁡t+…+ancos⁡nt+bnsin⁡nt)f(t) = e^{\alpha t}\left(a_{0} + a_{1} \cos t + + b_{1} \sin t + \ldots + a_{n} \cos nt + b_{n} \sin nt\right), where α\alpha is a fixed real number.

Problem 20.9

Prove that the given set of functions forms an infinite-dimensional vector space:

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(1)

the set of all polynomials

(2)

the set of all trigonometric polynomials

(3)

the set of functions continuous on some segment.

Problem 20.10

Find the linear combination of the columns:

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(1)

3c1−13c23\mathbf{c}_{1} - \frac{1}{3} \mathbf{c}_{2}. c1=[−3],c2=[6]\mathbf{c}_{1} = \left[\begin{smallmatrix} -3 \end{smallmatrix}\right], \mathbf{c}_{2} = \left[\begin{smallmatrix} 6 \end{smallmatrix}\right].

(2)

−c29+c33+2c31- \mathbf{c}_{29} + \mathbf{c}_{33} + 2\mathbf{c}_{31}. c29=[32],c33=[21],c31=[1−1]\mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{33} = \left[\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right].

(3)

12c143−12c138\frac{1}{2} \mathbf{c}_{143} - \frac{1}{2} \mathbf{c}_{138}. c143=[112],c138=[7012]\mathbf{c}_{143} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{138} = \left[\begin{smallmatrix} 7 \\ 0 \\ 12 \end{smallmatrix}\right].

(4)

c206−3c198+2c199\mathbf{c}_{206} - 3\mathbf{c}_{198} + 2\mathbf{c}_{199}. c206=[2233],c198=[1113],c199=[3−572]\mathbf{c}_{206} = \left[\begin{smallmatrix} 2 \\ 2 \\ 3 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right].

Problem 20.11

Find the linear combination of the matrices 13A215−12A252+A253−A254\frac{1}{3} \mathbf{A}_{215} - \frac{1}{2} \mathbf{A}_{252} + \mathbf{A}_{253} - \mathbf{A}_{254}. A215=[06163−31−3−3],A252=[0−1210−2−220],A253=[011−10−1−110],A254=[035−302−5−20]\mathbf{A}_{215} = \left[\begin{smallmatrix} 0 & 6 & 1 \\ 6 & 3 & -3 \\ 1 & -3 & -3 \end{smallmatrix}\right], \mathbf{A}_{252} = \left[\begin{smallmatrix} 0 & -1 & 2 \\ 1 & 0 & -2 \\ -2 & 2 & 0 \end{smallmatrix}\right], \mathbf{A}_{253} = \left[\begin{smallmatrix} 0 & 1 & 1 \\ -1 & 0 & -1 \\ -1 & 1 & 0 \end{smallmatrix}\right], \mathbf{A}_{254} = \left[\begin{smallmatrix} 0 & 3 & 5 \\ -3 & 0 & 2 \\ -5 & -2 & 0 \end{smallmatrix}\right].

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Problem 20.12

Find the column x\mathbf{x} from the equation:

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(1)

c28+c29−2x=c32\mathbf{c}_{28} + \mathbf{c}_{29} - 2\mathrm{x} = \mathbf{c}_{32}. c28=[−16],c29=[32],c32=[2−2]\mathbf{c}_{28} = \left[\begin{smallmatrix} -1 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right].

(2)

c141+x2−c146+x3=c142\frac{\mathbf{c}_{141} + \mathrm{x}}{2} - \frac{\mathbf{c}_{146} + \mathrm{x}}{3} = \mathbf{c}_{142}. c141=[211],c146=[141],c142=[011]\mathbf{c}_{141} = \left[\begin{smallmatrix} 2 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{146} = \left[\begin{smallmatrix} 1 \\ 4 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{142} = \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \end{smallmatrix}\right].

(3)

3(c197+x)+2(c202−x)=4(c204−x)3\left(\mathbf{c}_{197} + \mathbf{x}\right) + 2\left(\mathbf{c}_{202} - \mathbf{x}\right) = 4\left(\mathbf{c}_{204} - \mathbf{x}\right). c197=[1122],c202=[1013],c204=[3333]\mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{204} = \left[\begin{smallmatrix} 3 \\ 3 \\ 3 \\ 3 \end{smallmatrix}\right].

Problem 20.13

Identify the linear dependencies among the given columns:

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(1)

c34,c35,c29\mathbf{c}_{34}, \mathbf{c}_{35}, \mathbf{c}_{29}. c34=[13],c35=[14],c29=[32]\mathbf{c}_{34} = \left[\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{35} = \left[\begin{smallmatrix} 1 \\ 4 \end{smallmatrix}\right], \mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right].

(2)

c84,c83,c120\mathbf{c}_{84}, \mathbf{c}_{83}, \mathbf{c}_{120}. c84=[431],c83=[123],c120=[2−1−5]\mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{120} = \left[\begin{smallmatrix} 2 \\ -1 \\ -5 \end{smallmatrix}\right].

(3)

c166,c198,c199,c201\mathbf{c}_{166}, \mathbf{c}_{198}, \mathbf{c}_{199}, \mathbf{c}_{201}. c166=[1111],c198=[1113],c199=[3−572],c201=[1−75−2]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{201} = \left[\begin{smallmatrix} 1 \\ -7 \\ 5 \\ -2 \end{smallmatrix}\right].

(4)

c166,c197,c205,c206\mathbf{c}_{166}, \mathbf{c}_{197}, \mathbf{c}_{205}, \mathbf{c}_{206}. c166=[1111],c197=[1122],c205=[0011],c206=[2233]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{205} = \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{206} = \left[\begin{smallmatrix} 2 \\ 2 \\ 3 \\ 3 \end{smallmatrix}\right].

Problem 20.14

Find the dimension and a basis of the linear span of the given system of columns:

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(1)

c1,c2\mathbf{c}_{1}, \mathbf{c}_{2}. c1=[−3],c2=[6]\mathbf{c}_{1} = \left[\begin{smallmatrix} -3 \end{smallmatrix}\right], \mathbf{c}_{2} = \left[\begin{smallmatrix} 6 \end{smallmatrix}\right].

(2)

c31,c28,c30\mathbf{c}_{31}, \mathbf{c}_{28}, \mathbf{c}_{30}. c31=[1−1],c28=[−16],c30=[−11]\mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{28} = \left[\begin{smallmatrix} -1 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right].

(3)

c31,c30,c32\mathbf{c}_{31}, \mathbf{c}_{30}, \mathbf{c}_{32}. c31=[1−1],c30=[−11],c32=[2−2]\mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right].

(4)

c121,c124,c118\mathbf{c}_{121}, \mathbf{c}_{124}, \mathbf{c}_{118}. c121=[−320],c124=[−36−15],c118=[0−415]\mathbf{c}_{121} = \left[\begin{smallmatrix} -3 \\ 2 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{124} = \left[\begin{smallmatrix} -3 \\ 6 \\ -15 \end{smallmatrix}\right], \mathbf{c}_{118} = \left[\begin{smallmatrix} 0 \\ -4 \\ 15 \end{smallmatrix}\right].

(5)

c166,c198,c199,c201\mathbf{c}_{166}, \mathbf{c}_{198}, \mathbf{c}_{199}, \mathbf{c}_{201}. c166=[1111],c198=[1113],c199=[3−572],c201=[1−75−2]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{201} = \left[\begin{smallmatrix} 1 \\ -7 \\ 5 \\ -2 \end{smallmatrix}\right].

(6)

c196,c198,c202\mathbf{c}_{196}, \mathbf{c}_{198}, \mathbf{c}_{202}. c196=[1213],c198=[1113],c202=[1013]\mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right].

(7)

c166,c196,c197,c198\mathbf{c}_{166}, \mathbf{c}_{196}, \mathbf{c}_{197}, \mathbf{c}_{198}. c166=[1111],c196=[1213],c197=[1122],c198=[1113]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right].

(8)

o\mathbf{o}

(9)

c166,c203,c204,c197\mathbf{c}_{166}, \mathbf{c}_{203}, \mathbf{c}_{204}, \mathbf{c}_{197}. c166=[1111],c203=[2222],c204=[3333],c197=[1122]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{203} = \left[\begin{smallmatrix} 2 \\ 2 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{204} = \left[\begin{smallmatrix} 3 \\ 3 \\ 3 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right].

Problem 20.15

Find the dimension and a basis of the linear span of the system of matrices A391,A390,A389\mathbf{A}_{391}, \mathbf{A}_{390}, \mathbf{A}_{389}. A391=[689016],A390=[211301],A389=[276−614]\mathbf{A}_{391} = \left[\begin{smallmatrix} 6 & 8 & 9 \\ 0 & 1 & 6 \end{smallmatrix}\right], \mathbf{A}_{390} = \left[\begin{smallmatrix} 2 & 1 & 1 \\ 3 & 0 & 1 \end{smallmatrix}\right], \mathbf{A}_{389} = \left[\begin{smallmatrix} 2 & 7 & 6 \\ -6 & 1 & 4 \end{smallmatrix}\right].

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Problem 20.16

Find the dimension and a basis of the linear span of the system of polynomials (1+t)3,t3,1,t+t2(1 + t)^{3}, t^{3}, 1, t + t^{2}.

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Problem 20.17

Prove that the vectors e1,…,en\mathbf{e}_{1}, \ldots , \mathbf{e}_{n} form a basis in the nn-dimensional arithmetic space, and find the coordinate column of the vector x\mathbf{x} in this basis:

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(1)

n=1,e1=c1,x=c2n = 1, \mathbf{e}_{1} = \mathbf{c}_{1}, \mathbf{x} = \mathbf{c}_{2}. c1=[−3],c2=[6]\mathbf{c}_{1} = \left[\begin{smallmatrix} -3 \end{smallmatrix}\right], \mathbf{c}_{2} = \left[\begin{smallmatrix} 6 \end{smallmatrix}\right].

(2)

n=2,e1=c28,e2=c29,x=c30n = 2, \mathbf{e}_{1} = \mathbf{c}_{28}, \mathbf{e}_{2} = \mathbf{c}_{29}, \mathbf{x} = \mathbf{c}_{30}. c28=[−16],c29=[32],c30=[−11]\mathbf{c}_{28} = \left[\begin{smallmatrix} -1 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right].

(3)

n=3,e1=c116,e2=c120,e3=c122,x=c49n = 3, \mathbf{e}_{1} = \mathbf{c}_{116}, \mathbf{e}_{2} = \mathbf{c}_{120}, \mathbf{e}_{3} = \mathbf{c}_{122}, \mathbf{x} = \mathbf{c}_{49}. c116=[421],c120=[2−1−5],c122=[−231],c49=[2110]\mathbf{c}_{116} = \left[\begin{smallmatrix} 4 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{120} = \left[\begin{smallmatrix} 2 \\ -1 \\ -5 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{49} = \left[\begin{smallmatrix} 2 \\ 1 \\ 10 \end{smallmatrix}\right].

(4)

n=4,e1=c196,e2=c197,e3=c198,e4=c199,x=c200n = 4, \mathbf{e}_{1} = \mathbf{c}_{196}, \mathbf{e}_{2} = \mathbf{c}_{197}, \mathbf{e}_{3} = \mathbf{c}_{198}, \mathbf{e}_{4} = \mathbf{c}_{199}, \mathbf{x} = \mathbf{c}_{200}. c196=[1213],c197=[1122],c198=[1113],c199=[3−572],c200=[−18−65]\mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{200} = \left[\begin{smallmatrix} -1 \\ 8 \\ -6 \\ 5 \end{smallmatrix}\right].

(5)

n=5,e1=c255,e2=c263,e3=c264,e4=c265,e5=c266n = 5, \mathbf{e}_{1} = \mathbf{c}_{255}, \mathbf{e}_{2} = \mathbf{c}_{263}, \mathbf{e}_{3} = \mathbf{c}_{264}, \mathbf{e}_{4} = \mathbf{c}_{265}, \mathbf{e}_{5} = \mathbf{c}_{266}, x=c267\mathbf{x} = \mathbf{c}_{267}. c255=[11111],c263=[−12103],c264=[21421],c265=[02−153],c266=[700−10],c267=[44−1−26]\mathbf{c}_{255} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{263} = \left[\begin{smallmatrix} -1 \\ 2 \\ 1 \\ 0 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{264} = \left[\begin{smallmatrix} 2 \\ 1 \\ 4 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{265} = \left[\begin{smallmatrix} 0 \\ 2 \\ -1 \\ 5 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{266} = \left[\begin{smallmatrix} 7 \\ 0 \\ 0 \\ -1 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{267} = \left[\begin{smallmatrix} 4 \\ 4 \\ -1 \\ -2 \\ 6 \end{smallmatrix}\right].

Problem 20.18

Prove that the matrices A5,A10,A13,A6\mathbf{A}_{5}, \mathbf{A}_{10}, \mathbf{A}_{13}, \mathbf{A}_{6} form a basis in the space of square matrices of order 2, and find the coordinate column of the matrix A26\mathbf{A}_{26} in this basis. A5=[1−11−1],A10=[2513],A13=[1101],A6=[3457],A26=[514613]\mathbf{A}_{5} = \left[\begin{smallmatrix} 1 & -1 \\ 1 & -1 \end{smallmatrix}\right], \mathbf{A}_{10} = \left[\begin{smallmatrix} 2 & 5 \\ 1 & 3 \end{smallmatrix}\right], \mathbf{A}_{13} = \left[\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right], \mathbf{A}_{6} = \left[\begin{smallmatrix} 3 & 4 \\ 5 & 7 \end{smallmatrix}\right], \mathbf{A}_{26} = \left[\begin{smallmatrix} 5 & 14 \\ 6 & 13 \end{smallmatrix}\right].

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Problem 20.19

Prove that the matrices A200,A202,A205,A204,A203\mathbf{A}_{200}, \mathbf{A}_{202}, \mathbf{A}_{205}, \mathbf{A}_{204}, \mathbf{A}_{203}, A242\mathbf{A}_{242} form a basis in the space of symmetric matrices of order 3, and find the coordinate column of the matrix A215\mathbf{A}_{215} in this basis. A200=[001010100],A202=[12221−22−21],A205=[123234345],A204=[−231362121],A203=[22−12−12−122],A242=[0111−1−11−1−2],A215=[06163−31−3−3]\mathbf{A}_{200} = \left[\begin{smallmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{smallmatrix}\right], \mathbf{A}_{202} = \left[\begin{smallmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{smallmatrix}\right], \mathbf{A}_{205} = \left[\begin{smallmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{smallmatrix}\right], \mathbf{A}_{204} = \left[\begin{smallmatrix} -2 & 3 & 1 \\ 3 & 6 & 2 \\ 1 & 2 & 1 \end{smallmatrix}\right], \mathbf{A}_{203} = \left[\begin{smallmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ -1 & 2 & 2 \end{smallmatrix}\right], \mathbf{A}_{242} = \left[\begin{smallmatrix} 0 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -2 \end{smallmatrix}\right], \mathbf{A}_{215} = \left[\begin{smallmatrix} 0 & 6 & 1 \\ 6 & 3 & -3 \\ 1 & -3 & -3 \end{smallmatrix}\right].

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Problem 20.20

Prove that the polynomials 1,t−α,(t−α)2,……,(t−α)n1, t - \alpha , (t - \alpha )^{2}, \ldots \ldots , (t - \alpha )^{n} form a basis in the space of polynomials of degree at most nn, and find the coordinate column of an arbitrary polynomial pn(t)p_{n}(t) of degree at most nn in this basis.

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Problem 20.21

Prove that the polynomials 2t+t5,t3−t5,t+t32t + t^{5}, t^{3} - t^{5}, t + t^{3} form a basis in the space of odd polynomials of degree at most 5, and find the coordinate column of the polynomial 5t−t3+2t55t - t^{3} + 2t^{5} in this basis.

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Problem 20.22

Find the dimension and a basis of the vector subspace given in some basis by a system of linear equations:

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(1)

A27x=o\mathbf{A}_{27} \mathbf{x} = \mathbf{o}. A27=[1−3−26]\mathbf{A}_{27} = \left[\begin{smallmatrix} 1 & -3 \\ -2 & 6 \end{smallmatrix}\right].

(2)

A238x=0\mathbf{A}_{238} \mathbf{x} = \mathbf{0}. A238=[−310−10−74−4−2−33]\mathbf{A}_{238} = \left[\begin{smallmatrix} -3 & 10 & -10 \\ -7 & 4 & -4 \\ -2 & -3 & 3 \end{smallmatrix}\right].

(3)

A249x=o\mathbf{A}_{249} \mathrm{x} = \mathbf{o}. A249=[−31−26−24−15510]\mathbf{A}_{249} = \left[\begin{smallmatrix} -3 & 1 & -2 \\ 6 & -2 & 4 \\ -15 & 5 & 10 \end{smallmatrix}\right].

(4)

A391x=o\mathbf{A}_{391} \mathbf{x} = \mathbf{o}. A391=[689016]\mathbf{A}_{391} = \left[\begin{smallmatrix} 6 & 8 & 9 \\ 0 & 1 & 6 \end{smallmatrix}\right].

(5)

A110x=o\mathbf{A}_{110} \mathbf{x} = \mathbf{o}. A110=[312110]\mathbf{A}_{110} = \left[\begin{smallmatrix} 3 & 1 \\ 2 & 1 \\ 1 & 0 \end{smallmatrix}\right].

(6)

A442x=o\mathbf{A}_{442} \mathrm{x} = \mathbf{o}. A442=[33−4−3061154212332]\mathbf{A}_{442} = \left[\begin{smallmatrix} 3 & 3 & -4 & -3 \\ 0 & 6 & 1 & 1 \\ 5 & 4 & 2 & 1 \\ 2 & 3 & 3 & 2 \end{smallmatrix}\right].

(7)

A577x=o\mathbf{A}_{577} \mathrm{x} = \mathrm{o}. A577=[2−13−244−25172−1182]\mathbf{A}_{577} = \left[\begin{smallmatrix} 2 & -1 & 3 & -2 & 4 \\ 4 & -2 & 5 & 1 & 7 \\ 2 & -1 & 1 & 8 & 2 \end{smallmatrix}\right].

Problem 20.23

Compose a system of equations defining the linear span of the given system of columns:

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(1)

c66,c83\mathbf{c}_{66}, \mathbf{c}_{83}. c66=[111],c83=[123]\mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right].

(2)

c31,c30\mathbf{c}_{31}, \mathbf{c}_{30}. c31=[1−1],c30=[−11]\mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right].

(3)

c30,c29\mathbf{c}_{30}, \mathbf{c}_{29}. c30=[−11],c29=[32]\mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right].

(4)

c166,c196\mathbf{c}_{166}, \mathbf{c}_{196}. c166=[1111],c196=[1213]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right].

(5)

c197\mathbf{c}_{197}. c197=[1122]\mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right].

(6)

c166,c198,c199,c201\mathbf{c}_{166}, \mathbf{c}_{198}, \mathbf{c}_{199}, \mathbf{c}_{201}. c166=[1111],c198=[1113],c199=[3−572],c201=[1−75−2]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{201} = \left[\begin{smallmatrix} 1 \\ -7 \\ 5 \\ -2 \end{smallmatrix}\right].

(7)

c166,c196,c197,c198\mathbf{c}_{166}, \mathbf{c}_{196}, \mathbf{c}_{197}, \mathbf{c}_{198}. c166=[1111],c196=[1213],c197=[1122],c198=[1113]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right].

(8)

[0000]\left[\begin{smallmatrix} 0\\ 0\\ 0\\ 0\end{smallmatrix}\right].

Problem 20.24

Prove that each of the two systems of vectors f1,……,fnf_{1}, \ldots \ldots , \mathbf{f}_{n} and g1,…,gn\mathbf{g}_{1}, \ldots , \mathbf{g}_{n} is a basis in the nn-dimensional arithmetic space, and find the transition matrix from the first basis to the second. Find the coordinates of a vector in the first basis, if its coordinates ξ1′,…,ξn′\xi_{1}^{\prime }, \ldots , \xi_{n}^{\prime } in the second basis are known:

?
(1)

n=1,f1=c1,g1=c3n = 1, \quad \mathbf{f}_{1} = \mathbf{c}_{1}, \quad \mathbf{g}_{1} = \mathbf{c}_{3}. c1=[−3],c3=[−9]\mathbf{c}_{1} = \left[\begin{smallmatrix} -3 \end{smallmatrix}\right], \mathbf{c}_{3} = \left[\begin{smallmatrix} -9 \end{smallmatrix}\right].

(2)

n=2,f1=c29,f2=c33,g1=c32,g2=c28n = 2, \quad \mathbf{f}_{1} = \mathbf{c}_{29}, \quad \mathbf{f}_{2} = \mathbf{c}_{33}, \quad \mathbf{g}_{1} = \mathbf{c}_{32}, \quad \mathbf{g}_{2} = \mathbf{c}_{28}. c29=[32],c33=[21],c32=[2−2],c28=[−16]\mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{33} = \left[\begin{smallmatrix} 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{28} = \left[\begin{smallmatrix} -1 \\ 6 \end{smallmatrix}\right].

(3)

n=3,f1=c116,f2=c117,f3=c94,g1=c119,g2=c84n = 3, \mathbf{f}_{1} = \mathbf{c}_{116}, \mathbf{f}_{2} = \mathbf{c}_{117}, \mathbf{f}_{3} = \mathbf{c}_{94}, \mathbf{g}_{1} = \mathbf{c}_{119}, \mathbf{g}_{2} = \mathbf{c}_{84}, g3=c83\mathbf{g}_{3} = \mathbf{c}_{83}. c116=[421],c117=[532],c94=[321],c119=[−140],c84=[431],c83=[123]\mathbf{c}_{116} = \left[\begin{smallmatrix} 4 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{117} = \left[\begin{smallmatrix} 5 \\ 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{94} = \left[\begin{smallmatrix} 3 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{119} = \left[\begin{smallmatrix} -1 \\ 4 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right].

(4)

n=4,f1=c166,f2=c196,f3=c197,f4=c198,g1=c199n = 4, \mathbf{f}_{1} = \mathbf{c}_{166}, \mathbf{f}_{2} = \mathbf{c}_{196}, \mathbf{f}_{3} = \mathbf{c}_{197}, \mathbf{f}_{4} = \mathbf{c}_{198}, \mathbf{g}_{1} = \mathbf{c}_{199}, g2=c200,g3=c202,g4=c203\mathbf{g}_{2} = \mathbf{c}_{200}, \mathbf{g}_{3} = \mathbf{c}_{202}, \mathbf{g}_{4} = \mathbf{c}_{203}. c166=[1111],c196=[1213],c197=[1122],c198=[1113],c199=[3−572],c200=[−18−65],c202=[1013],c203=[2222]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{200} = \left[\begin{smallmatrix} -1 \\ 8 \\ -6 \\ 5 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{203} = \left[\begin{smallmatrix} 2 \\ 2 \\ 2 \\ 2 \end{smallmatrix}\right].

Problem 20.25

Prove that each of the two systems of matrices A132\mathbf{A}_{132}, A143,A134,A133,A110,A135\mathbf{A}_{143}, \mathbf{A}_{134}, \mathbf{A}_{133}, \mathbf{A}_{110}, \mathbf{A}_{135} and A136,A137,A112,A138,A139,A113\mathbf{A}_{136}, \mathbf{A}_{137}, \mathbf{A}_{112}, \mathbf{A}_{138}, \mathbf{A}_{139}, \mathbf{A}_{113} is a basis in the space of matrices of size 3×23 \times 2, and find the transition matrix from the first basis to the second. Find the coordinates of a 3×23 \times 2 matrix in the first basis, if its coordinates ξ1′,…,ξ6′\xi_{1}^{\prime }, \ldots , \xi_{6}^{\prime } in the second basis are known. A132=[123002],A143=[01−1100],A134=[001−205],A133=[000107],A110=[312110],A135=[000001],A136=[113500],A137=[214023],A112=[001−110],A138=[339000],A139=[000111],A113=[010110]\mathbf{A}_{132} = \left[\begin{smallmatrix} 1 & 2 \\ 3 & 0 \\ 0 & 2 \end{smallmatrix}\right], \mathbf{A}_{143} = \left[\begin{smallmatrix} 0 & 1 \\ -1 & 1 \\ 0 & 0 \end{smallmatrix}\right], \mathbf{A}_{134} = \left[\begin{smallmatrix} 0 & 0 \\ 1 & -2 \\ 0 & 5 \end{smallmatrix}\right], \mathbf{A}_{133} = \left[\begin{smallmatrix} 0 & 0 \\ 0 & 1 \\ 0 & 7 \end{smallmatrix}\right], \mathbf{A}_{110} = \left[\begin{smallmatrix} 3 & 1 \\ 2 & 1 \\ 1 & 0 \end{smallmatrix}\right], \mathbf{A}_{135} = \left[\begin{smallmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{smallmatrix}\right], \mathbf{A}_{136} = \left[\begin{smallmatrix} 1 & 1 \\ 3 & 5 \\ 0 & 0 \end{smallmatrix}\right], \mathbf{A}_{137} = \left[\begin{smallmatrix} 2 & 1 \\ 4 & 0 \\ 2 & 3 \end{smallmatrix}\right], \mathbf{A}_{112} = \left[\begin{smallmatrix} 0 & 0 \\ 1 & -1 \\ 1 & 0 \end{smallmatrix}\right], \mathbf{A}_{138} = \left[\begin{smallmatrix} 3 & 3 \\ 9 & 0 \\ 0 & 0 \end{smallmatrix}\right], \mathbf{A}_{139} = \left[\begin{smallmatrix} 0 & 0 \\ 0 & 1 \\ 1 & 1 \end{smallmatrix}\right], \mathbf{A}_{113} = \left[\begin{smallmatrix} 0 & 1 \\ 0 & 1 \\ 1 & 0 \end{smallmatrix}\right].

?
Problem 20.26

Prove that each of the two systems of matrices A250,A251,A252\mathbf{A}_{250}, \mathbf{A}_{251}, \mathbf{A}_{252} and A253,A254,A255\mathbf{A}_{253}, \mathbf{A}_{254}, \mathbf{A}_{255} is a basis in the space of skew-symmetric matrices of order 3, and find the transition matrix from the first basis to the second. Find the coordinates of a skew-symmetric matrix of order 3 in the first basis, if its coordinates ξ1′,ξ2′,ξ3′\xi_{1}^{\prime }, \xi_{2}^{\prime }, \xi_{3}^{\prime } in the second basis are known. A250=[01−2−1032−30],A251=[00−10041−40],A252=[0−1210−2−220],A253=[011−10−1−110],A254=[035−302−5−20],A255=[010−1030−30]\mathbf{A}_{250} = \left[\begin{smallmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{smallmatrix}\right], \mathbf{A}_{251} = \left[\begin{smallmatrix} 0 & 0 & -1 \\ 0 & 0 & 4 \\ 1 & -4 & 0 \end{smallmatrix}\right], \mathbf{A}_{252} = \left[\begin{smallmatrix} 0 & -1 & 2 \\ 1 & 0 & -2 \\ -2 & 2 & 0 \end{smallmatrix}\right], \mathbf{A}_{253} = \left[\begin{smallmatrix} 0 & 1 & 1 \\ -1 & 0 & -1 \\ -1 & 1 & 0 \end{smallmatrix}\right], \mathbf{A}_{254} = \left[\begin{smallmatrix} 0 & 3 & 5 \\ -3 & 0 & 2 \\ -5 & -2 & 0 \end{smallmatrix}\right], \mathbf{A}_{255} = \left[\begin{smallmatrix} 0 & 1 & 0 \\ -1 & 0 & 3 \\ 0 & -3 & 0 \end{smallmatrix}\right].

?
Problem 20.27

Prove that each of the two systems of functions t−t2,t3,1+5t+t3,(1+t)3t - t^{2}, t^{3}, 1 + 5t + t^{3}, (1 + t)^{3} and (1+t)3,(1−t)3,t−t2+t3,1+t+t2+t3(1 + t)^{3}, (1 - t)^{3}, t - t^{2} + t^{3}, 1 + t + t^{2} + t^{3} is a basis in the space of polynomials of degree at most 3. Find the transition matrix from the first basis to the second and the coordinates of a polynomial in the first basis, if its coordinates ξ1′,ξ2′,ξ3′,ξ4′\xi_{1}^{\prime }, \xi_{2}^{\prime }, \xi_{3}^{\prime }, \xi_{4}^{\prime } in the second basis are known.

?
Problem 20.28

Prove that each of the two systems of functions (1+t2)2,(1−t2)2,1\left(1 + t^{2}\right)^{2}, \left(1 - t^{2}\right)^{2}, 1 and 1+t2+t4,1−t2+t4,t41 + t^{2} + t^{4}, 1 - t^{2} + t^{4}, t^{4} is a basis in the space of even polynomials of degree at most 4, and find the transition matrix from the first basis to the second. Find the coordinates of an even polynomial of degree at most 4 in the first basis, if its coordinates ξ1′,ξ2′,ξ3′\xi_{1}^{\prime }, \xi_{2}^{\prime }, \xi_{3}^{\prime } in the second basis are known.

?
Problem 20.29

How will the transition matrix from one basis to another change if:

?
(1)

the ii-th and jj-th vectors of the first basis are interchanged

(2)

the ii-th and jj-th vectors of the second basis are interchanged

(3)

the vectors of both bases are arranged in reverse order.

Problem 20.30

The matrix S1S_{1} is the transition matrix from the first basis e1,…,ene_{1}, \ldots , e_{n} in an nn-dimensional vector space to the second basis f1,…,fnf_{1}, \ldots , f_{n}, and the matrix S2S_{2} is the transition matrix from the second basis to the third basis g1,…,gng_{1}, \ldots , g_{n}. Find the transition matrix:

?
(1)

from the first basis to the third

(2)

from the second basis to the first.

Problem 20.31

Describe the mutual arrangement of two bases in a three-dimensional vector space, if the transition matrix from the first basis to the second is:

?
(1)

diagonal

(2)

upper triangular

(3)

lower triangular.

Problem 20.32

The basis vectors a1,…,ana_{1}, \ldots , a_{n} and b1,…,bnb_{1}, \ldots , b_{n} are given by their coordinate columns with respect to basis e. These columns form, respectively, the matrices AA and BB. Find the transition matrix from basis a\mathbf{a} to basis b\mathbf{b}, if:

?
(1)

A=A54,B=A64A = \mathbf{A}_{54}, \quad B = \mathbf{A}_{64}. A54=[1/103/10−3/101/10],A64=[3/101/101/10−3/10]\mathbf{A}_{54} = \left[\begin{smallmatrix} 1/\sqrt{10} & 3/\sqrt{10} \\ -3/\sqrt{10} & 1/\sqrt{10} \end{smallmatrix}\right], \mathbf{A}_{64} = \left[\begin{smallmatrix} 3/\sqrt{10} & 1/\sqrt{10} \\ 1/\sqrt{10} & -3/\sqrt{10} \end{smallmatrix}\right].

(2)

A=A10,B=A34A = \mathbf{A}_{10}, \quad B = \mathbf{A}_{34}. A10=[2513],A34=[2111]\mathbf{A}_{10} = \left[\begin{smallmatrix} 2 & 5 \\ 1 & 3 \end{smallmatrix}\right], \mathbf{A}_{34} = \left[\begin{smallmatrix} 2 & 1 \\ 1 & 1 \end{smallmatrix}\right].

(3)

A=A427,B=A468A = \mathbf{A}_{427}, B = \mathbf{A}_{468}. A427=[1331131−111−13−1−133],A468=[111−111−111−111−1111]\mathbf{A}_{427} = \left[\begin{smallmatrix} 1 & 3 & 3 & 1 \\ 1 & 3 & 1 & -1 \\ 1 & 1 & -1 & 3 \\ -1 & -1 & 3 & 3 \end{smallmatrix}\right], \mathbf{A}_{468} = \left[\begin{smallmatrix} 1 & 1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 1 & -1 & 1 & 1 \\ -1 & 1 & 1 & 1 \end{smallmatrix}\right].

Problem 20.33

Prove that the Legendre polynomials 1,t1, t, 12(3t2−1),12(5t3−3t)\frac{1}{2}\left(3t^{2} - 1\right), \frac{1}{2}\left(5t^{3} - 3t\right) form a basis in the space of polynomials of degree at most 3. Find the transition matrices from the standard basis to the given one and back.

?
Problem 20.34

Find the coordinates of the polynomial p(t)p(t) in the standard basis of the space of polynomials of degree at most 3 and in the basis of problem 20. 33:

?
(1)

p(t)=5t3−3tp(t) = 5t^{3} - 3t; \quad

(2)

p(t)=2t−1p(t) = 2t - 1

(3)

p(t)=9t2−1p(t) = 9t^{2} - 1

(4)

p(t)=1−4t−3t2+10t3p(t) = 1 - 4t - 3t^{2} + 10t^{3}.

Section 21Sum and intersection of subspaces[17/35%]
Problem 21.1

Prove that the space of square matrices of order nn is the direct sum of the subspace of symmetric matrices and the subspace of skew-symmetric matrices of the same order.

?
Problem 21.2

Prove that the space of polynomials of degree at most nn is the direct sum of the subspace of even polynomials of degree at most nn and the subspace of odd polynomials of degree at most nn.

?
Problem 21.3
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(1)

Prove that an nn-dimensional vector space is the direct sum of the subspace of vectors all of whose coordinates are equal to each other, and the subspace of vectors the sum of whose coordinates equals 0.

(2)

A matrix AA of nn rows is given. Prove that the nn-dimensional arithmetic space is the direct sum of the linear span of the columns of AA and the subspace of solutions of the system of linear equations ATx=0A^{T} \mathbf{x} = \mathbf{0}.

Problem 21.4

Prove that the nn-dimensional arithmetic space is the direct sum of the vector subspaces spanned by the systems of vectors a1,…,ak\mathbf{a}_{1}, \ldots , \mathbf{a}_{k} and b1,…,bl\mathbf{b}_{1}, \ldots , \mathbf{b}_{l}:

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(1)

n=2,a1=c30,a2=c32,b1=c35n = 2, \quad \mathbf{a}_{1} = \mathbf{c}_{30}, \quad \mathbf{a}_{2} = \mathbf{c}_{32}, \quad \mathbf{b}_{1} = \mathbf{c}_{35}. c30=[−11],c32=[2−2],c35=[14]\mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{35} = \left[\begin{smallmatrix} 1 \\ 4 \end{smallmatrix}\right].

(2)

n=3,a1=c141,a2=c146,b1=c66,b2=c140n = 3, \quad \mathbf{a}_{1} = \mathbf{c}_{141}, \quad \mathbf{a}_{2} = \mathbf{c}_{146}, \quad \mathbf{b}_{1} = \mathbf{c}_{66}, \quad \mathbf{b}_{2} = \mathbf{c}_{140}. c141=[211],c146=[141],c66=[111],c140=[444]\mathbf{c}_{141} = \left[\begin{smallmatrix} 2 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{146} = \left[\begin{smallmatrix} 1 \\ 4 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{140} = \left[\begin{smallmatrix} 4 \\ 4 \\ 4 \end{smallmatrix}\right].

(3)

n=4,a1=c198,a2=c196,b1=c197,b2=c166,b3=c206n = 4, \mathbf{a}_{1} = \mathbf{c}_{198}, \mathbf{a}_{2} = \mathbf{c}_{196}, \mathbf{b}_{1} = \mathbf{c}_{197}, \mathbf{b}_{2} = \mathbf{c}_{166}, \mathbf{b}_{3} = \mathbf{c}_{206}. c198=[1113],c196=[1213],c197=[1122],c166=[1111],c206=[2233]\mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{206} = \left[\begin{smallmatrix} 2 \\ 2 \\ 3 \\ 3 \end{smallmatrix}\right].

(4)

n=4,a1=c196,a2=c198,a3=c202,a4=c199,b1=c166n = 4, \mathbf{a}_{1} = \mathbf{c}_{196}, \mathbf{a}_{2} = \mathbf{c}_{198}, \mathbf{a}_{3} = \mathbf{c}_{202}, \mathbf{a}_{4} = \mathbf{c}_{199}, \mathbf{b}_{1} = \mathbf{c}_{166}, b2=c204\mathbf{b}_{2} = \mathbf{c}_{204}. c196=[1213],c198=[1113],c202=[1013],c199=[3−572],c166=[1111],c204=[3333]\mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{204} = \left[\begin{smallmatrix} 3 \\ 3 \\ 3 \\ 3 \end{smallmatrix}\right].

Problem 21.5

Decompose the given vector x\mathbf{x} of the nn-dimensional arithmetic space into a sum of two vectors, one of which lies in P\mathcal{P}, and the other in Q\mathcal{Q}, where P\mathcal{P} is the linear span of the system of vectors a1,…,ak\mathbf{a}_{1}, \ldots , \mathbf{a}_{k}, and Q\mathcal{Q} is the linear span of the system of vectors b1,…,bl\mathbf{b}_{1}, \ldots , \mathbf{b}_{l}. Verify the uniqueness of the decomposition:

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(1)

n=2,x=c29,a1=c28,b1=c30n = 2, \quad \mathbf{x} = \mathbf{c}_{29}, \quad \mathbf{a}_{1} = \mathbf{c}_{28}, \quad \mathbf{b}_{1} = \mathbf{c}_{30}. c29=[32],c28=[−16],c30=[−11]\mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{28} = \left[\begin{smallmatrix} -1 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right].

(2)

n=3,x=c120,a1=c84,a2=c83,b1=c66n = 3, \quad \mathbf{x} = \mathbf{c}_{120}, \quad \mathbf{a}_{1} = \mathbf{c}_{84}, \quad \mathbf{a}_{2} = \mathbf{c}_{83}, \quad \mathbf{b}_{1} = \mathbf{c}_{66}. c120=[2−1−5],c84=[431],c83=[123],c66=[111]\mathbf{c}_{120} = \left[\begin{smallmatrix} 2 \\ -1 \\ -5 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right].

(3)

n=3,x=c145,a1=c84,a2=c83,b1=c66n = 3, \quad \mathbf{x} = \mathbf{c}_{145}, \quad \mathbf{a}_{1} = \mathbf{c}_{84}, \quad \mathbf{a}_{2} = \mathbf{c}_{83}, \quad \mathbf{b}_{1} = \mathbf{c}_{66}. c145=[20−1],c84=[431],c83=[123],c66=[111]\mathbf{c}_{145} = \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right].

(4)

n=3,x=c139,a1=c84,a2=c83,b1=c66n = 3, \quad \mathbf{x} = \mathbf{c}_{139}, \quad \mathbf{a}_{1} = \mathbf{c}_{84}, \quad \mathbf{a}_{2} = \mathbf{c}_{83}, \quad \mathbf{b}_{1} = \mathbf{c}_{66}. c139=[−2−2−2],c84=[431],c83=[123],c66=[111]\mathbf{c}_{139} = \left[\begin{smallmatrix} -2 \\ -2 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right].

(5)

n=4,x=c200,a1=c166,a2=c198,a3=c207,b1=c202n = 4, \mathbf{x} = \mathbf{c}_{200}, \mathbf{a}_{1} = \mathbf{c}_{166}, \mathbf{a}_{2} = \mathbf{c}_{198}, \mathbf{a}_{3} = \mathbf{c}_{207}, \mathbf{b}_{1} = \mathbf{c}_{202}, b2=c205\mathbf{b}_{2} = \mathbf{c}_{205}. c200=[−18−65],c166=[1111],c198=[1113],c207=[1112],c202=[1013],c205=[0011]\mathbf{c}_{200} = \left[\begin{smallmatrix} -1 \\ 8 \\ -6 \\ 5 \end{smallmatrix}\right], \mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{207} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{205} = \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \\ 1 \end{smallmatrix}\right].

Problem 21.6

Find the projection of the given vector x\mathbf{x} of the nn-dimensional arithmetic space onto the vector subspace P\mathcal{P} parallel to the vector subspace Q\mathcal{Q}, where P\mathcal{P} is the linear span of the system of vectors a1,…,ak\mathbf{a}_{1}, \ldots , \mathbf{a}_{k}, and Q\mathcal{Q} is the linear span of the system of vectors b1,…,bl\mathbf{b}_{1}, \ldots , \mathbf{b}_{l}:

?
(1)

n=2,x=c32,a1=c30,b1=c34n = 2, \quad \mathbf{x} = \mathbf{c}_{32}, \quad \mathbf{a}_{1} = \mathbf{c}_{30}, \quad \mathbf{b}_{1} = \mathbf{c}_{34}. c32=[2−2],c30=[−11],c34=[13]\mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{34} = \left[\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}\right].

(2)

n=2,x=c37,a1=c30,b1=c34n = 2, \quad \mathbf{x} = \mathbf{c}_{37}, \quad \mathbf{a}_{1} = \mathbf{c}_{30}, \quad \mathbf{b}_{1} = \mathbf{c}_{34}. c37=[26],c30=[−11],c34=[13]\mathbf{c}_{37} = \left[\begin{smallmatrix} 2 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{34} = \left[\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}\right].

(3)

n=2,x=c35,a1=c30,b1=c34n = 2, \quad \mathbf{x} = \mathbf{c}_{35}, \quad \mathbf{a}_{1} = \mathbf{c}_{30}, \quad \mathbf{b}_{1} = \mathbf{c}_{34}. c35=[14],c30=[−11],c34=[13]\mathbf{c}_{35} = \left[\begin{smallmatrix} 1 \\ 4 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{34} = \left[\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}\right].

(4)

n=3,x=c146,a1=c66,a2=c121,a3=c122,b1=c145n = 3, \mathbf{x} = \mathbf{c}_{146}, \mathbf{a}_{1} = \mathbf{c}_{66}, \mathbf{a}_{2} = \mathbf{c}_{121}, \mathbf{a}_{3} = \mathbf{c}_{122}, \mathbf{b}_{1} = \mathbf{c}_{145}. c146=[141],c66=[111],c121=[−320],c122=[−231],c145=[20−1]\mathbf{c}_{146} = \left[\begin{smallmatrix} 1 \\ 4 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{121} = \left[\begin{smallmatrix} -3 \\ 2 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{145} = \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \end{smallmatrix}\right].

(5)

n=4,x=c201,a1=c166,a2=c199,b1=c197,b2=c198n = 4, \mathbf{x} = \mathbf{c}_{201}, \mathbf{a}_{1} = \mathbf{c}_{166}, \mathbf{a}_{2} = \mathbf{c}_{199}, \mathbf{b}_{1} = \mathbf{c}_{197}, \mathbf{b}_{2} = \mathbf{c}_{198}. c201=[1−75−2],c166=[1111],c199=[3−572],c197=[1122],c198=[1113]\mathbf{c}_{201} = \left[\begin{smallmatrix} 1 \\ -7 \\ 5 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{199} = \left[\begin{smallmatrix} 3 \\ -5 \\ 7 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right].

Problem 21.7

Find the dimension and a basis of the sum and intersection of the vector subspaces of the nn-dimensional arithmetic space spanned by the systems of vectors a1,…,ak\mathbf{a}_{1}, \ldots , \mathbf{a}_{k} and b1,…,bl\mathbf{b}_{1}, \ldots , \mathbf{b}_{l}:

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(1)

n=2,a1=c34,a2=c37,a3=c35,b1=c30,b2=c36n = 2, \mathbf{a}_{1} = \mathbf{c}_{34}, \mathbf{a}_{2} = \mathbf{c}_{37}, \mathbf{a}_{3} = \mathbf{c}_{35}, \mathbf{b}_{1} = \mathbf{c}_{30}, \mathbf{b}_{2} = \mathbf{c}_{36}, b3=c32\mathbf{b}_{3} = \mathbf{c}_{32}. c34=[13],c37=[26],c35=[14],c30=[−11],c36=[3−3],c32=[2−2]\mathbf{c}_{34} = \left[\begin{smallmatrix} 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{37} = \left[\begin{smallmatrix} 2 \\ 6 \end{smallmatrix}\right], \mathbf{c}_{35} = \left[\begin{smallmatrix} 1 \\ 4 \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{36} = \left[\begin{smallmatrix} 3 \\ -3 \end{smallmatrix}\right], \mathbf{c}_{32} = \left[\begin{smallmatrix} 2 \\ -2 \end{smallmatrix}\right].

(2)

n=3,a1=c116,a2=c121,a3=c119,b1=c122,b2=c125n = 3, \mathbf{a}_{1} = \mathbf{c}_{116}, \mathbf{a}_{2} = \mathbf{c}_{121}, \mathbf{a}_{3} = \mathbf{c}_{119}, \mathbf{b}_{1} = \mathbf{c}_{122}, \mathbf{b}_{2} = \mathbf{c}_{125}, b3=c138\mathbf{b}_{3} = \mathbf{c}_{138}. c116=[421],c121=[−320],c119=[−140],c122=[−231],c125=[5313],c138=[7012]\mathbf{c}_{116} = \left[\begin{smallmatrix} 4 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{121} = \left[\begin{smallmatrix} -3 \\ 2 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{119} = \left[\begin{smallmatrix} -1 \\ 4 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{125} = \left[\begin{smallmatrix} 5 \\ 3 \\ 13 \end{smallmatrix}\right], \mathbf{c}_{138} = \left[\begin{smallmatrix} 7 \\ 0 \\ 12 \end{smallmatrix}\right].

(3)

n=3,a1=c66,a2=c139,a3=c140,b1=c94,b2=c141n = 3, \mathbf{a}_{1} = \mathbf{c}_{66}, \mathbf{a}_{2} = \mathbf{c}_{139}, \mathbf{a}_{3} = \mathbf{c}_{140}, \mathbf{b}_{1} = \mathbf{c}_{94}, \mathbf{b}_{2} = \mathbf{c}_{141}, b3=c117\mathbf{b}_{3} = \mathbf{c}_{117}. c66=[111],c139=[−2−2−2],c140=[444],c94=[321],c141=[211],c117=[532]\mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{139} = \left[\begin{smallmatrix} -2 \\ -2 \\ -2 \end{smallmatrix}\right], \mathbf{c}_{140} = \left[\begin{smallmatrix} 4 \\ 4 \\ 4 \end{smallmatrix}\right], \mathbf{c}_{94} = \left[\begin{smallmatrix} 3 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{141} = \left[\begin{smallmatrix} 2 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{117} = \left[\begin{smallmatrix} 5 \\ 3 \\ 2 \end{smallmatrix}\right].

(4)

(р) n=3,a1=c83,a2=c142,a3=c143,b1=c84,b2=c144,b3=c117n = 3, \mathbf{a}_{1} = \mathbf{c}_{83}, \mathbf{a}_{2} = \mathbf{c}_{142}, \mathbf{a}_{3} = \mathbf{c}_{143}, \mathbf{b}_{1} = \mathbf{c}_{84}, \mathbf{b}_{2} = \mathbf{c}_{144}, \mathbf{b}_{3} = \mathbf{c}_{117}. c83=[123],c142=[011],c143=[112],c84=[431],c144=[110],c117=[532]\mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{142} = \left[\begin{smallmatrix} 0 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{143} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{144} = \left[\begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{117} = \left[\begin{smallmatrix} 5 \\ 3 \\ 2 \end{smallmatrix}\right].

(5)

n=3,a1=c66,a2=c116,a3=c145,b1=c122,b2=c146n = 3, \mathbf{a}_{1} = \mathbf{c}_{66}, \mathbf{a}_{2} = \mathbf{c}_{116}, \mathbf{a}_{3} = \mathbf{c}_{145}, \mathbf{b}_{1} = \mathbf{c}_{122}, \mathbf{b}_{2} = \mathbf{c}_{146}, b3=c147\mathbf{b}_{3} = \mathbf{c}_{147}. c66=[111],c116=[421],c145=[20−1],c122=[−231],c146=[141],c147=[5−2−1]\mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{116} = \left[\begin{smallmatrix} 4 \\ 2 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{145} = \left[\begin{smallmatrix} 2 \\ 0 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{146} = \left[\begin{smallmatrix} 1 \\ 4 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{147} = \left[\begin{smallmatrix} 5 \\ -2 \\ -1 \end{smallmatrix}\right].

(6)

n=3,a1=c83,a2=c84,a3=c120,b1=c66,b2=c121n = 3, \mathbf{a}_{1} = \mathbf{c}_{83}, \mathbf{a}_{2} = \mathbf{c}_{84}, \mathbf{a}_{3} = \mathbf{c}_{120}, \mathbf{b}_{1} = \mathbf{c}_{66}, \mathbf{b}_{2} = \mathbf{c}_{121}, b3=c122\mathbf{b}_{3} = \mathbf{c}_{122}. c83=[123],c84=[431],c120=[2−1−5],c66=[111],c121=[−320],c122=[−231]\mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{84} = \left[\begin{smallmatrix} 4 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{120} = \left[\begin{smallmatrix} 2 \\ -1 \\ -5 \end{smallmatrix}\right], \mathbf{c}_{66} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{121} = \left[\begin{smallmatrix} -3 \\ 2 \\ 0 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right].

(7)

n=4,a1=c196,a2=c200,a3=c217,b1=c211,b2=c218n = 4, \mathbf{a}_{1} = \mathbf{c}_{196}, \mathbf{a}_{2} = \mathbf{c}_{200}, \mathbf{a}_{3} = \mathbf{c}_{217}, \mathbf{b}_{1} = \mathbf{c}_{211}, \mathbf{b}_{2} = \mathbf{c}_{218}, b3=c219\mathbf{b}_{3} = \mathbf{c}_{219}. c196=[1213],c200=[−18−65],c217=[010−58],c211=[14−15],c218=[3−263],c219=[4258]\mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{200} = \left[\begin{smallmatrix} -1 \\ 8 \\ -6 \\ 5 \end{smallmatrix}\right], \mathbf{c}_{217} = \left[\begin{smallmatrix} 0 \\ 10 \\ -5 \\ 8 \end{smallmatrix}\right], \mathbf{c}_{211} = \left[\begin{smallmatrix} 1 \\ 4 \\ -1 \\ 5 \end{smallmatrix}\right], \mathbf{c}_{218} = \left[\begin{smallmatrix} 3 \\ -2 \\ 6 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{219} = \left[\begin{smallmatrix} 4 \\ 2 \\ 5 \\ 8 \end{smallmatrix}\right].

(8)

n=4,a1=c196,a2=c198,a3=c202,b1=c166,b2=c204n = 4, \mathbf{a}_{1} = \mathbf{c}_{196}, \mathbf{a}_{2} = \mathbf{c}_{198}, \mathbf{a}_{3} = \mathbf{c}_{202}, \mathbf{b}_{1} = \mathbf{c}_{166}, \mathbf{b}_{2} = \mathbf{c}_{204}, b3=c197\mathbf{b}_{3} = \mathbf{c}_{197}. c196=[1213],c198=[1113],c202=[1013],c166=[1111],c204=[3333],c197=[1122]\mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{204} = \left[\begin{smallmatrix} 3 \\ 3 \\ 3 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right].

(9)

n=4,a1=c166,a2=c196,a3=c197,b1=c203,b2=c205n = 4, \mathbf{a}_{1} = \mathbf{c}_{166}, \mathbf{a}_{2} = \mathbf{c}_{196}, \mathbf{a}_{3} = \mathbf{c}_{197}, \mathbf{b}_{1} = \mathbf{c}_{203}, \mathbf{b}_{2} = \mathbf{c}_{205}, b3=c206\mathbf{b}_{3} = \mathbf{c}_{206}. c166=[1111],c196=[1213],c197=[1122],c203=[2222],c205=[0011],c206=[2233]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{203} = \left[\begin{smallmatrix} 2 \\ 2 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{205} = \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{206} = \left[\begin{smallmatrix} 2 \\ 2 \\ 3 \\ 3 \end{smallmatrix}\right].

(10)

n=4,a1=c166,a2=c198,a3=c196,a4=c202,b1=c207,b2=c204,b3=c197,b4=c205n = 4, \mathbf{a}_{1} = \mathbf{c}_{166}, \mathbf{a}_{2} = \mathbf{c}_{198}, \mathbf{a}_{3} = \mathbf{c}_{196}, \mathbf{a}_{4} = \mathbf{c}_{202}, \mathbf{b}_{1} = \mathbf{c}_{207}, \mathbf{b}_{2} = \mathbf{c}_{204}, \mathbf{b}_{3} = \mathbf{c}_{197}, \mathbf{b}_{4} = \mathbf{c}_{205}. c166=[1111],c198=[1113],c196=[1213],c202=[1013],c207=[1112],c204=[3333],c197=[1122],c205=[0011]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{198} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{202} = \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{207} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{204} = \left[\begin{smallmatrix} 3 \\ 3 \\ 3 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{197} = \left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{205} = \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \\ 1 \end{smallmatrix}\right].

(11)

(р) n=4,a1=[1−110],a2=[1100]n = 4, \quad \mathbf{a}_{1} =\left[\begin{smallmatrix} 1 \\ - 1 \\ 1 \\ 0\end{smallmatrix}\right], \quad \mathbf{a}_{2} =\left[\begin{smallmatrix} 1 \\ 1 \\ 0 \\ 0\end{smallmatrix}\right], a3=[2012],a4=[2011],b1=[4614]\mathbf{a}_{3} =\left[\begin{smallmatrix} 2 \\ 0 \\ 1 \\ 2\end{smallmatrix}\right], \quad \mathbf{a}_{4} =\left[\begin{smallmatrix} 2 \\ 0 \\ 1 \\ 1\end{smallmatrix}\right], \quad \mathbf{b}_{1} =\left[\begin{smallmatrix} 4 \\ 6 \\ 1 \\ 4\end{smallmatrix}\right], b2=[1120],b3=[11−23],b4=[2411]\mathbf{b}_{2} =\left[\begin{smallmatrix} 1 \\ 1 \\ 2 \\ 0\end{smallmatrix}\right], \quad \mathbf{b}_{3} =\left[\begin{smallmatrix} 1 \\ 1 \\ - 2 \\ 3\end{smallmatrix}\right], \quad \mathbf{b}_{4} =\left[\begin{smallmatrix} 2 \\ 4 \\ 1 \\ 1\end{smallmatrix}\right].

Problem 21.8

Find the dimension and a basis of the sum and intersection of the vector subspaces of the space of square matrices of order 3, spanned by the systems of matrices A202,A201,A209,A204\mathbf{A}_{202}, \mathbf{A}_{201}, \mathbf{A}_{209}, \mathbf{A}_{204} and A256,A205,A257,A258\mathbf{A}_{256}, \mathbf{A}_{205}, \mathbf{A}_{257}, \mathbf{A}_{258}. A202=[12221−22−21],A201=[21−131−2101],A209=[1−3−1−27232−4],A204=[−231362121],A256=[001026164],A205=[123234345],A257=[0544−42029],A258=[0534−6−4−1−45]\mathbf{A}_{202} = \left[\begin{smallmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{smallmatrix}\right], \mathbf{A}_{201} = \left[\begin{smallmatrix} 2 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & 0 & 1 \end{smallmatrix}\right], \mathbf{A}_{209} = \left[\begin{smallmatrix} 1 & -3 & -1 \\ -2 & 7 & 2 \\ 3 & 2 & -4 \end{smallmatrix}\right], \mathbf{A}_{204} = \left[\begin{smallmatrix} -2 & 3 & 1 \\ 3 & 6 & 2 \\ 1 & 2 & 1 \end{smallmatrix}\right], \mathbf{A}_{256} = \left[\begin{smallmatrix} 0 & 0 & 1 \\ 0 & 2 & 6 \\ 1 & 6 & 4 \end{smallmatrix}\right], \mathbf{A}_{205} = \left[\begin{smallmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{smallmatrix}\right], \mathbf{A}_{257} = \left[\begin{smallmatrix} 0 & 5 & 4 \\ 4 & -4 & 2 \\ 0 & 2 & 9 \end{smallmatrix}\right], \mathbf{A}_{258} = \left[\begin{smallmatrix} 0 & 5 & 3 \\ 4 & -6 & -4 \\ -1 & -4 & 5 \end{smallmatrix}\right].

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Problem 21.9

Find the dimension and a basis of the sum and intersection of the vector subspaces of the space of polynomials of degree at most 3, spanned by the systems of polynomials 1+2t+t31 + 2t + t^{3}, 1+t+t2,t−t2+t31 + t + t^{2}, t - t^{2} + t^{3} and 1+t2,1+3t+t3,3t−t2+t31 + t^{2}, 1 + 3t + t^{3}, 3t - t^{2} + t^{3}.

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Problem 21.10

Using the notion of the sum of two vector subspaces, prove the inequality rg⁡(A+B)⩽rg⁡A+rg⁡B\operatorname {rg}(A + B) \leqslant \operatorname {rg} A + \operatorname {rg} B, where AA and BB are two matrices of the same size m×nm \times n.

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Problem 21.11

Prove that the sum L\mathcal{L} of two vector subspaces P\mathcal{P} and Q\mathcal{Q} is a direct sum if and only if at least one vector x∈Lx\in \mathcal{L} is uniquely represented in the form x=y+zx = y + z, where y∈P,z∈Qy\in \mathcal{P}, z\in \mathcal{Q}.

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Problem 21.12

Let P\mathcal{P} and Q\mathcal{Q} be two vector subspaces of a finite-dimensional vector space. Prove that:

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(1)

if the sum of the dimensions of P\mathcal{P} and Q\mathcal{Q} is greater than the dimension of the whole space, then the intersection P∩Q\mathcal{P} \cap \mathcal{Q} contains a nonzero vector

(2)

if the dimension of the sum of P\mathcal{P} and Q\mathcal{Q} is one more than the dimension of their intersection, then one of these subspaces is contained in the other.

Problem 21.13

Prove that for any vector subspace P\mathcal{P} of a finite-dimensional vector space there exists another subspace Q\mathcal{Q} such that the whole space is the direct sum of P\mathcal{P} and Q\mathcal{Q}.

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Problem 21.14

Let L,M,N\mathcal{L}, \mathcal{M}, \mathcal{N} be three vector subspaces; P=(L∩N)+(M∩N),Q=(L+M)∩N\mathcal{P} = (\mathcal{L} \cap \mathcal{N}) + (\mathcal{M} \cap \mathcal{N}), \mathcal{Q} = (\mathcal{L} + \mathcal{M}) \cap \mathcal{N}

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(1)

Prove that P⊆Q\mathcal{P} \subseteq \mathcal{Q}.

(2)

Is the case P≠Q\mathcal{P} \neq \mathcal{Q} possible? Give a corresponding example.

Problem 21.15

Prove that the sum of the vector subspaces L(1),…,L(k)\mathcal{L}^{(1)}, \ldots , \mathcal{L}^{(k)} coincides with the set of vectors representable in the form x=x1+…+xkx = x_{1} + \ldots + x_{k}, where xi∈L(i),i=1,…,kx_{i} \in \mathcal{L}^{(i)}, i = 1, \ldots , k.

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Problem 21.16

Let L,M,N\mathcal{L}, \mathcal{M}, \mathcal{N} be three vector subspaces. Prove that

L+M+N=(L+M)+N=L+(M+N) \mathcal{L} + \mathcal{M} + \mathcal{N} = (\mathcal{L} + \mathcal{M}) + \mathcal{N} = \mathcal{L} + (\mathcal{M} + \mathcal{N})
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Problem 21.17

Prove that the sum L\mathcal{L} of the vector subspaces L(1),…,L(k)\mathcal{L}^{(1)}, \ldots , \mathcal{L}^{(k)} is a direct sum if and only if at least one vector x∈Lx\in \mathcal{L} is uniquely represented in the form x=x1+…+xkx = x_{1} + \ldots + x_{k}, where xi∈L(i)x_{i} \in \mathcal{L}^{(i)} (a generalization of problem 21. 11).

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Section 22Complex linear spaces[13/100%]
Problem 22.1

Find the linear combination of the columns:

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(1)

(1+i)c4−ic5(1 + i) \mathbf{c}_{4} - i\mathbf{c}_{5}. c4=[1−2i],c5=[7+i]\mathbf{c}_{4} = \left[\begin{smallmatrix} 1-2i \end{smallmatrix}\right], \mathbf{c}_{5} = \left[\begin{smallmatrix} 7+i \end{smallmatrix}\right].

(2)

2ic42+(3+i)c30−c412i\mathbf{c}_{42} + (3 + i) \mathbf{c}_{30} - \mathbf{c}_{41}. c42=[3−i3−6i],c30=[−11],c41=[1+2i6+2i]\mathbf{c}_{42} = \left[\begin{smallmatrix} 3-i \\ 3-6 i \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{41} = \left[\begin{smallmatrix} 1+2 i \\ 6+2 i \end{smallmatrix}\right].

(3)

(1−2i)c148+32c152(1 - 2i) \mathbf{c}_{148} + \frac{3}{2} \mathbf{c}_{152}. c148=[203+i],c152=[3i−6i−3]\mathbf{c}_{148} = \left[\begin{smallmatrix} 2 \\ 0 \\ 3+i \end{smallmatrix}\right], \mathbf{c}_{152} = \left[\begin{smallmatrix} 3 i \\ -6 i \\ -3 \end{smallmatrix}\right].

Problem 22.2

Find the linear combination of the matrices (2+i)A89+iA10+A27−2A91(2 + i) \mathbf{A}_{89} + i\mathbf{A}_{10} + \mathbf{A}_{27} - 2\mathbf{A}_{91}. A89=[1−i2+i1−2i],A10=[2513],A27=[1−3−26],A91=[1−i2+i6−4i9+7i]\mathbf{A}_{89} = \left[\begin{smallmatrix} 1 & -i \\ 2+i & 1-2 i \end{smallmatrix}\right], \mathbf{A}_{10} = \left[\begin{smallmatrix} 2 & 5 \\ 1 & 3 \end{smallmatrix}\right], \mathbf{A}_{27} = \left[\begin{smallmatrix} 1 & -3 \\ -2 & 6 \end{smallmatrix}\right], \mathbf{A}_{91} = \left[\begin{smallmatrix} 1-i & 2+i \\ 6-4 i & 9+7 i \end{smallmatrix}\right].

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Problem 22.3

Find the column x\mathbf{x} from the equation:

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(1)

(1+i)(x−c44)−(2+i)(x+c45)=c43(1 + i)\left(\mathbf{x} - \mathbf{c}_{44}\right) - (2 + i)\left(\mathbf{x} + \mathbf{c}_{45}\right) = \mathbf{c}_{43}. c44=[i2],c45=[2+3i−3i],c43=[1−i−2−2i]\mathbf{c}_{44} = \left[\begin{smallmatrix} i \\ 2 \end{smallmatrix}\right], \mathbf{c}_{45} = \left[\begin{smallmatrix} 2+3 i \\ -3 i \end{smallmatrix}\right], \mathbf{c}_{43} = \left[\begin{smallmatrix} 1-i \\ -2-2i \end{smallmatrix}\right].

(2)

2c153−c149+ix=c1502\mathbf{c}_{153} - \mathbf{c}_{149} + i\mathbf{x} = \mathbf{c}_{150}. c153=[−i−1−3i],c149=[143+2i],c150=[−143−4i]\mathbf{c}_{153} = \left[\begin{smallmatrix} -i \\ -1 \\ -3 i \end{smallmatrix}\right], \mathbf{c}_{149} = \left[\begin{smallmatrix} 1 \\ 4 \\ 3+2 i \end{smallmatrix}\right], \mathbf{c}_{150} = \left[\begin{smallmatrix} -1 \\ 4 \\ 3-4 i \end{smallmatrix}\right].

Problem 22.4

Identify the linear dependencies among the given columns:

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(1)

c26,c29,c43\mathbf{c}_{26}, \mathbf{c}_{29}, \mathbf{c}_{43}. c26=[−12i],c29=[32],c43=[1−i−2−2i]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{29} = \left[\begin{smallmatrix} 3 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{43} = \left[\begin{smallmatrix} 1-i \\ -2-2i \end{smallmatrix}\right].

(2)

c26,c30,c40\mathbf{c}_{26}, \mathbf{c}_{30}, \mathbf{c}_{40}. c26=[−12i],c30=[−11],c40=[1+i3]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{30} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{40} = \left[\begin{smallmatrix} 1+i \\ 3 \end{smallmatrix}\right].

(3)

c131,c132,c133\mathbf{c}_{131}, \mathbf{c}_{132}, \mathbf{c}_{133}. c131=[1−i1+i],c132=[103i],c133=[−12i−2+i]\mathbf{c}_{131} = \left[\begin{smallmatrix} 1 \\ -i \\ 1+i \end{smallmatrix}\right], \mathbf{c}_{132} = \left[\begin{smallmatrix} 1 \\ 0 \\ 3 i \end{smallmatrix}\right], \mathbf{c}_{133} = \left[\begin{smallmatrix} -1 \\ 2 i \\ -2+i \end{smallmatrix}\right].

Problem 22.5

Find the dimension and a basis of the linear span of the given system of columns:

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(1)

c5\mathbf{c}_{5}. c5=[7+i]\mathbf{c}_{5} = \left[\begin{smallmatrix} 7+i \end{smallmatrix}\right].

(2)

c27,c39\mathbf{c}_{27}, \mathbf{c}_{39}. c27=[−3+2i−i],c39=[3+i7−6i]\mathbf{c}_{27} = \left[\begin{smallmatrix} -3+2i \\ -i \end{smallmatrix}\right], \mathbf{c}_{39} = \left[\begin{smallmatrix} 3 + i \\ 7 - 6i \end{smallmatrix}\right].

(3)

c26,c43,c44\mathbf{c}_{26}, \mathbf{c}_{43}, \mathbf{c}_{44}. c26=[−12i],c43=[1−i−2−2i],c44=[i2]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{43} = \left[\begin{smallmatrix} 1-i \\ -2-2i \end{smallmatrix}\right], \mathbf{c}_{44} = \left[\begin{smallmatrix} i \\ 2 \end{smallmatrix}\right].

(4)

c134,c151,c152\mathbf{c}_{134}, \mathbf{c}_{151}, \mathbf{c}_{152}. c134=[1−2i],c151=[2−i−4+2i1+2i],c152=[3i−6i−3]\mathbf{c}_{134} = \left[\begin{smallmatrix} 1 \\ -2 \\ i \end{smallmatrix}\right], \mathbf{c}_{151} = \left[\begin{smallmatrix} 2-i \\ -4+2 i \\ 1+2 i \end{smallmatrix}\right], \mathbf{c}_{152} = \left[\begin{smallmatrix} 3 i \\ -6 i \\ -3 \end{smallmatrix}\right].

(5)

c275,c215,c276\mathbf{c}_{275}, \mathbf{c}_{215}, \mathbf{c}_{276}. c275=[03+i4−i−3],c215=[1111−i],c276=[17+2i9−2i−5−i]\mathbf{c}_{275} = \left[\begin{smallmatrix} 0 \\ 3+i \\ 4-i \\ -3 \end{smallmatrix}\right], \mathbf{c}_{215} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 - i \end{smallmatrix}\right], \mathbf{c}_{276} = \left[\begin{smallmatrix} 1 \\ 7+2 i \\ 9-2 i \\ -5-i \end{smallmatrix}\right].

(6)

c166,c215,c196,c193,c216\mathbf{c}_{166}, \mathbf{c}_{215}, \mathbf{c}_{196}, \mathbf{c}_{193}, \mathbf{c}_{216}. c166=[1111],c215=[1111−i],c196=[1213],c193=[0001],c216=[2324−i]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{215} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 - i \end{smallmatrix}\right], \mathbf{c}_{196} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{193} = \left[\begin{smallmatrix} 0 \\ 0 \\ 0 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{216} = \left[\begin{smallmatrix} 2 \\ 3 \\ 2 \\ 4 - i \end{smallmatrix}\right].

Problem 22.6

Prove that the vectors e1,…,en\mathbf{e}_{1}, \ldots , \mathbf{e}_{n} form a basis in the complex nn-dimensional arithmetic space, and find the coordinate column of the vector x\mathbf{x} in this basis:

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(1)

n=1,e1=c4,x=c5n = 1, \mathbf{e}_{1} = \mathbf{c}_{4}, \quad \mathbf{x} = \mathbf{c}_{5}. c4=[1−2i],c5=[7+i]\mathbf{c}_{4} = \left[\begin{smallmatrix} 1-2i \end{smallmatrix}\right], \mathbf{c}_{5} = \left[\begin{smallmatrix} 7+i \end{smallmatrix}\right].

(2)

n=2,e1=c31,e2=c35,x=c39n = 2, \quad \mathbf{e}_{1} = \mathbf{c}_{31}, \quad \mathbf{e}_{2} = \mathbf{c}_{35}, \quad \mathbf{x} = \mathbf{c}_{39}. c31=[1−1],c35=[14],c39=[3+i7−6i]\mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{35} = \left[\begin{smallmatrix} 1 \\ 4 \end{smallmatrix}\right], \mathbf{c}_{39} = \left[\begin{smallmatrix} 3 + i \\ 7 - 6i \end{smallmatrix}\right].

(3)

n=2,e1=c26,e2=c40,x=c41n = 2, \quad \mathbf{e}_{1} = \mathbf{c}_{26}, \quad \mathbf{e}_{2} = \mathbf{c}_{40}, \mathbf{x} = \mathbf{c}_{41}. c26=[−12i],c40=[1+i3],c41=[1+2i6+2i]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{40} = \left[\begin{smallmatrix} 1+i \\ 3 \end{smallmatrix}\right], \mathbf{c}_{41} = \left[\begin{smallmatrix} 1+2 i \\ 6+2 i \end{smallmatrix}\right].

(4)

n=2,e1=c26,e2=c27,x=c38n = 2, \quad \mathbf{e}_{1} = \mathbf{c}_{26}, \quad \mathbf{e}_{2} = \mathbf{c}_{27}, \quad \mathbf{x} = \mathbf{c}_{38}. c26=[−12i],c27=[−3+2i−i],c38=[5+8i−5+2i]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{27} = \left[\begin{smallmatrix} -3+2i \\ -i \end{smallmatrix}\right], \mathbf{c}_{38} = \left[\begin{smallmatrix} 5+8 i \\ -5+2 i \end{smallmatrix}\right].

(5)

n=3,e1=c126,e2=c127,e3=c128,x=c273n = 3, \mathbf{e}_{1} = \mathbf{c}_{126}, \mathbf{e}_{2} = \mathbf{c}_{127}, \mathbf{e}_{3} = \mathbf{c}_{128}, \quad \mathbf{x} = \mathbf{c}_{273}. c126=[1−i2+i3],c127=[0−1+2i3−i],c128=[−21+i1],c273=[−3−i6+4i4−3i]\mathbf{c}_{126} = \left[\begin{smallmatrix} 1-i \\ 2+i \\ 3 \end{smallmatrix}\right], \mathbf{c}_{127} = \left[\begin{smallmatrix} 0 \\ -1+2 i \\ 3-i \end{smallmatrix}\right], \mathbf{c}_{128} = \left[\begin{smallmatrix} -2 \\ 1+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{273} = \left[\begin{smallmatrix} -3-i \\ 6+4 i \\ 4-3 i \end{smallmatrix}\right].

(6)

n=4,e1=c166,e2=c207,e3=c274,e4=c275,x=c276n = 4, \mathbf{e}_{1} = \mathbf{c}_{166}, \mathbf{e}_{2} = \mathbf{c}_{207}, \mathbf{e}_{3} = \mathbf{c}_{274}, \mathbf{e}_{4} = \mathbf{c}_{275}, \mathbf{x} = \mathbf{c}_{276}. c166=[1111],c207=[1112],c274=[i1+i−3+2i0],c275=[03+i4−i−3],c276=[17+2i9−2i−5−i]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{207} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 2 \end{smallmatrix}\right], \mathbf{c}_{274} = \left[\begin{smallmatrix} i \\ 1+i \\ -3+2 i \\ 0 \end{smallmatrix}\right], \mathbf{c}_{275} = \left[\begin{smallmatrix} 0 \\ 3+i \\ 4-i \\ -3 \end{smallmatrix}\right], \mathbf{c}_{276} = \left[\begin{smallmatrix} 1 \\ 7+2 i \\ 9-2 i \\ -5-i \end{smallmatrix}\right].

Problem 22.7

Find the dimension and a basis of the vector space given in some basis by a system of linear equations:

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(1)

A90x=0\mathbf{A}_{90} \mathrm{x} = \mathbf{0}. A90=[1+i1+3i1−2i1+2i]\mathbf{A}_{90} = \left[\begin{smallmatrix} 1+i & 1+3 i \\ 1-2 i & 1+2 i \end{smallmatrix}\right].

(2)

A91x=o\mathbf{A}_{91} \mathrm{x} = \mathbf{o}. A91=[1−i2+i6−4i9+7i]\mathbf{A}_{91} = \left[\begin{smallmatrix} 1-i & 2+i \\ 6-4 i & 9+7 i \end{smallmatrix}\right].

(3)

A369x=o\mathbf{A}_{369} \mathbf{x} = \mathbf{o}. A369=[1−i−3+2i2−i−4+6i4−3i−3i−9+i5−i4]\mathbf{A}_{369} = \left[\begin{smallmatrix} 1-i & -3+2 i & 2-i \\ -4+6 i & 4-3 i & -3 i \\ -9+i & 5-i & 4 \end{smallmatrix}\right].

(4)

A371x=o\mathbf{A}_{371} \mathbf{x} = \mathbf{o}. A371=[11−i2+i1−3i−2−4i5−5i2i2+2i−2+4i]\mathbf{A}_{371} = \left[\begin{smallmatrix} 1 & 1-i & 2+i \\ 1-3 i & -2-4 i & 5-5 i \\ 2 i & 2+2 i & -2+4 i \end{smallmatrix}\right].

(5)

A525x=o\mathbf{A}_{525} \mathrm{x} = \mathrm{o}. A525=[1−21+2i−1+i−13−1−3i2−ii1+i12i2i02i−2+2i02−i−1−2i2−i]\mathbf{A}_{525} = \left[\begin{smallmatrix} 1 & -2 & 1+2 i & -1+i \\ -1 & 3 & -1-3 i & 2-i \\ i & 1+i & 1 & 2 i \\ 2 i & 0 & 2 i & -2+2 i \\ 0 & 2-i & -1-2 i & 2-i \end{smallmatrix}\right].

Problem 22.8

Compose a system of equations defining the linear span of the given system of columns:

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(1)

c40,c42\mathbf{c}_{40}, \mathbf{c}_{42}. c40=[1+i3],c42=[3−i3−6i]\mathbf{c}_{40} = \left[\begin{smallmatrix} 1+i \\ 3 \end{smallmatrix}\right], \mathbf{c}_{42} = \left[\begin{smallmatrix} 3-i \\ 3-6 i \end{smallmatrix}\right].

(2)

c26,c42\mathbf{c}_{26}, \mathbf{c}_{42}. c26=[−12i],c42=[3−i3−6i]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right], \mathbf{c}_{42} = \left[\begin{smallmatrix} 3-i \\ 3-6 i \end{smallmatrix}\right].

(3)

c215\mathbf{c}_{215}. c215=[1111−i]\mathbf{c}_{215} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 - i \end{smallmatrix}\right].

(4)

c128,c273\mathbf{c}_{128}, \mathbf{c}_{273}. c128=[−21+i1],c273=[−3−i6+4i4−3i]\mathbf{c}_{128} = \left[\begin{smallmatrix} -2 \\ 1+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{273} = \left[\begin{smallmatrix} -3-i \\ 6+4 i \\ 4-3 i \end{smallmatrix}\right].

(5)

c275,c276,c214,c215\mathbf{c}_{275}, \mathbf{c}_{276}, \mathbf{c}_{214}, \mathbf{c}_{215}. c275=[03+i4−i−3],c276=[17+2i9−2i−5−i],c214=[14+i5−i−2−i],c215=[1111−i]\mathbf{c}_{275} = \left[\begin{smallmatrix} 0 \\ 3+i \\ 4-i \\ -3 \end{smallmatrix}\right], \mathbf{c}_{276} = \left[\begin{smallmatrix} 1 \\ 7+2 i \\ 9-2 i \\ -5-i \end{smallmatrix}\right], \mathbf{c}_{214} = \left[\begin{smallmatrix} 1 \\ 4+i \\ 5-i \\ -2-i \end{smallmatrix}\right], \mathbf{c}_{215} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 - i \end{smallmatrix}\right].

Problem 22.9

Prove that each of the two systems of vectors f1,……,fnf_{1}, \ldots \ldots , \mathbf{f}_{n} and g1,…,gn\mathbf{g}_{1}, \ldots , \mathbf{g}_{n} is a basis in the complex nn-dimensional arithmetic space, and find the transition matrix from the first basis to the second. Find the coordinates of a vector in the first basis, if its coordinates ξ1′,…,ξn′\xi_{1}^{\prime }, \ldots , \xi_{n}^{\prime } in the second basis are known:

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(1)

n=1,f1=c4,g1=c6n = 1, \mathbf{f}_{1} = \mathbf{c}_{4}, \mathbf{g}_{1} = \mathbf{c}_{6}. c4=[1−2i],c6=[6−7i]\mathbf{c}_{4} = \left[\begin{smallmatrix} 1-2i \end{smallmatrix}\right], \mathbf{c}_{6} = \left[\begin{smallmatrix} 6-7 i \end{smallmatrix}\right].

(2)

n=2,f1=c31,f2=c45,g1=c44,g2=c39n = 2, \mathbf{f}_{1} = \mathbf{c}_{31}, \mathbf{f}_{2} = \mathbf{c}_{45}, \mathbf{g}_{1} = \mathbf{c}_{44}, \mathbf{g}_{2} = \mathbf{c}_{39}. c31=[1−1],c45=[2+3i−3i],c44=[i2],c39=[3+i7−6i]\mathbf{c}_{31} = \left[\begin{smallmatrix} 1 \\ -1 \end{smallmatrix}\right], \mathbf{c}_{45} = \left[\begin{smallmatrix} 2+3 i \\ -3 i \end{smallmatrix}\right], \mathbf{c}_{44} = \left[\begin{smallmatrix} i \\ 2 \end{smallmatrix}\right], \mathbf{c}_{39} = \left[\begin{smallmatrix} 3 + i \\ 7 - 6i \end{smallmatrix}\right].

(3)

n=3,f1=c129,f2=c128,f3=c130, g1=c122, g2=c126n = 3, f_{1} = \mathbf{c}_{129}, \mathbf{f}_{2} = \mathbf{c}_{128}, \mathbf{f}_{3} = \mathbf{c}_{130}, \mathrm{~ g}_{1} = \mathbf{c}_{122}, \mathrm{~ g}_{2} = \mathbf{c}_{126}, g3=c94\mathbf{g}_{3} = \mathbf{c}_{94}. c129=[−12+i1],c128=[−21+i1],c130=[−2+i1+i1],c122=[−231],c126=[1−i2+i3],c94=[321]\mathbf{c}_{129} = \left[\begin{smallmatrix} -1 \\ 2+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{128} = \left[\begin{smallmatrix} -2 \\ 1+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{130} = \left[\begin{smallmatrix} -2+i \\ 1+i \\ 1 \end{smallmatrix}\right], \mathbf{c}_{122} = \left[\begin{smallmatrix} -2 \\ 3 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{126} = \left[\begin{smallmatrix} 1-i \\ 2+i \\ 3 \end{smallmatrix}\right], \mathbf{c}_{94} = \left[\begin{smallmatrix} 3 \\ 2 \\ 1 \end{smallmatrix}\right].

Problem 22.10

Find the projection of the given vector x\mathbf{x} of the two-dimensional complex arithmetic space onto the vector subspace P\mathcal{P} parallel to the vector subspace Q\mathcal{Q}, where P\mathcal{P} is spanned by the vector c44\mathbf{c}_{44}, and Q\mathcal{Q} is spanned by the vector c40\mathbf{c}_{40}. c44=[i2],c40=[1+i3]\mathbf{c}_{44} = \left[\begin{smallmatrix} i \\ 2 \end{smallmatrix}\right], \mathbf{c}_{40} = \left[\begin{smallmatrix} 1+i \\ 3 \end{smallmatrix}\right].

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(1)

x=c26\mathbf{x} = \mathbf{c}_{26}. c26=[−12i]\mathbf{c}_{26} = \left[\begin{smallmatrix} -1 \\ 2i \end{smallmatrix}\right].

(2)

x=c42\mathbf{x} = \mathbf{c}_{42}. c42=[3−i3−6i]\mathbf{c}_{42} = \left[\begin{smallmatrix} 3-i \\ 3-6 i \end{smallmatrix}\right].

(3)

x=c45\mathbf{x} = \mathbf{c}_{45}. c45=[2+3i−3i]\mathbf{c}_{45} = \left[\begin{smallmatrix} 2+3 i \\ -3 i \end{smallmatrix}\right].

Problem 22.11

Find the dimension and a basis of the sum and intersection of the vector subspaces of the complex nn-dimensional arithmetic space spanned by the systems of vectors a1,……,ak\mathbf{a}_{1}, \ldots \ldots , \mathbf{a}_{k} and b1,…,bl\mathbf{b}_{1}, \ldots , \mathbf{b}_{l}:

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(1)

n=3,a1=c83,a2=c134,a3=c148,b1=c132,b2=c149n = 3, \mathbf{a}_{1} = \mathbf{c}_{83}, \mathbf{a}_{2} = \mathbf{c}_{134}, \mathbf{a}_{3} = \mathbf{c}_{148}, \mathbf{b}_{1} = \mathbf{c}_{132}, \mathbf{b}_{2} = \mathbf{c}_{149}, b3=c150\mathbf{b}_{3} = \mathbf{c}_{150}. c83=[123],c134=[1−2i],c148=[203+i],c132=[103i],c149=[143+2i],c150=[−143−4i]\mathbf{c}_{83} = \left[\begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}\right], \mathbf{c}_{134} = \left[\begin{smallmatrix} 1 \\ -2 \\ i \end{smallmatrix}\right], \mathbf{c}_{148} = \left[\begin{smallmatrix} 2 \\ 0 \\ 3+i \end{smallmatrix}\right], \mathbf{c}_{132} = \left[\begin{smallmatrix} 1 \\ 0 \\ 3 i \end{smallmatrix}\right], \mathbf{c}_{149} = \left[\begin{smallmatrix} 1 \\ 4 \\ 3+2 i \end{smallmatrix}\right], \mathbf{c}_{150} = \left[\begin{smallmatrix} -1 \\ 4 \\ 3-4 i \end{smallmatrix}\right].

(2)

n=3,a1=c131,a2=c132,a3=c133,b1=c134,b2=c135n = 3, \mathbf{a}_{1} = \mathbf{c}_{131}, \mathbf{a}_{2} = \mathbf{c}_{132}, \mathbf{a}_{3} = \mathbf{c}_{133}, \mathbf{b}_{1} = \mathbf{c}_{134}, \mathbf{b}_{2} = \mathbf{c}_{135}, b3=c136\mathbf{b}_{3} = \mathbf{c}_{136}. c131=[1−i1+i],c132=[103i],c133=[−12i−2+i],c134=[1−2i],c135=[21+i−i],c136=[05+i−3i]\mathbf{c}_{131} = \left[\begin{smallmatrix} 1 \\ -i \\ 1+i \end{smallmatrix}\right], \mathbf{c}_{132} = \left[\begin{smallmatrix} 1 \\ 0 \\ 3 i \end{smallmatrix}\right], \mathbf{c}_{133} = \left[\begin{smallmatrix} -1 \\ 2 i \\ -2+i \end{smallmatrix}\right], \mathbf{c}_{134} = \left[\begin{smallmatrix} 1 \\ -2 \\ i \end{smallmatrix}\right], \mathbf{c}_{135} = \left[\begin{smallmatrix} 2 \\ 1+i \\ -i \end{smallmatrix}\right], \mathbf{c}_{136} = \left[\begin{smallmatrix} 0 \\ 5+i \\ -3 i \end{smallmatrix}\right].

(3)

n=4,a1=c166,a2=c220,a3=c216,a4=c215,b1=c221n = 4, \mathbf{a}_{1} = \mathbf{c}_{166}, \mathbf{a}_{2} = \mathbf{c}_{220}, \mathbf{a}_{3} = \mathbf{c}_{216}, \mathbf{a}_{4} = \mathbf{c}_{215}, \mathbf{b}_{1} = \mathbf{c}_{221}, b2=c222,b3=c223,b4=c214\mathbf{b}_{2} = \mathbf{c}_{222}, \mathbf{b}_{3} = \mathbf{c}_{223}, \mathbf{b}_{4} = \mathbf{c}_{214}. c166=[1111],c220=[1213−i],c216=[2324−i],c215=[1111−i],c221=[0103−i],c222=[0205−2i],c223=[02+i06+i],c214=[14+i5−i−2−i]\mathbf{c}_{166} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 \end{smallmatrix}\right], \mathbf{c}_{220} = \left[\begin{smallmatrix} 1 \\ 2 \\ 1 \\ 3-i \end{smallmatrix}\right], \mathbf{c}_{216} = \left[\begin{smallmatrix} 2 \\ 3 \\ 2 \\ 4 - i \end{smallmatrix}\right], \mathbf{c}_{215} = \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \\ 1 - i \end{smallmatrix}\right], \mathbf{c}_{221} = \left[\begin{smallmatrix} 0 \\ 1 \\ 0 \\ 3-i \end{smallmatrix}\right], \mathbf{c}_{222} = \left[\begin{smallmatrix} 0 \\ 2 \\ 0 \\ 5-2 i \end{smallmatrix}\right], \mathbf{c}_{223} = \left[\begin{smallmatrix} 0 \\ 2+i \\ 0 \\ 6+i \end{smallmatrix}\right], \mathbf{c}_{214} = \left[\begin{smallmatrix} 1 \\ 4+i \\ 5-i \\ -2-i \end{smallmatrix}\right].

Problem 22.12
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(1)

Prove that if in an nn-dimensional complex vector space one considers multiplication of vectors only by real numbers, then one obtains a 2n2n-dimensional real vector space.

(2)

In the two-dimensional complex arithmetic space, the operation of multiplying vectors only by real numbers is considered. Find a basis in the resulting real space and the coordinate column of the vector c27\mathbf{c}_{27} in this basis. c27=[−3+2i−i]\mathbf{c}_{27} = \left[\begin{smallmatrix} -3+2i \\ -i \end{smallmatrix}\right].

Problem 22.13

Prove that the set of polynomials of degree at most nn with complex coefficients can be considered both as a complex vector space and as a real vector space. In both cases find:

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(1)

a basis and the dimension of the space

(2)

the coordinate column of the polynomial 1−2i+(3+i)t−3t21 - 2i + (3 + i) t - 3t^{2} in the basis found (for n=2)n = 2).