Section 22

Complex linear spaces

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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).