Chapter 2

Linear Equations

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Problem 2.1.1

Let

A=(112−1),b→=(15). A = \begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix}, \qquad \overrightarrow {b} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}.

Find the solutions of the system of linear equations Ax→=b→A\overrightarrow {x} = \overrightarrow {b}.

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Problem 2.1.2

Let α,b1,b2∈R\alpha , b_{1}, b_{2} \in \mathbb {R}. Solve the system of linear equations

(cos⁡(α)−sin⁡(α)sin⁡(α)cos⁡(α))(x1x2)=(b1b2). \begin{pmatrix} \cos (\alpha ) & -\sin (\alpha ) \\ \sin (\alpha ) & \cos (\alpha ) \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix}.
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Problem 2.1.3
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(i)

Let ϵ∈R\epsilon \in \mathbb {R} and

A=(1122),b→=(3ϵ). A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}, \qquad \overrightarrow {b} = \begin{pmatrix} 3 \\ \epsilon \end{pmatrix}.

Find the condition on ϵ\epsilon so that there is a solution of Ax→=b→A\overrightarrow {x} = \overrightarrow {b}.

(ii)

Consider the system of three linear equations

(1111241410)(x1x2x3)=(1ϵϵ2) \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 \\ \epsilon \\ \epsilon ^{2} \end{pmatrix}

with ϵ∈R\epsilon \in \mathbb {R}. Find the condition on ϵ\epsilon so that there is a solution.

Problem 2.1.4

Find all solutions of the system of linear equations

(cos⁡(θ)−sin⁡(θ)−sin⁡(θ)−cos⁡(θ))(x1x2)=(x1x2),θ∈R \begin{pmatrix} \cos (\theta ) & -\sin (\theta ) \\ -\sin (\theta ) & -\cos (\theta ) \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}, \qquad \theta \in \mathbb {R}

with x→≠0→\overrightarrow {x} \neq \overrightarrow {0}. What type of equation is this?

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Problem 2.1.5

Find all solutions of the system of linear equations

(5−2−4−222−425)(x1x2x3)=(x1x2x3). \begin{pmatrix} 5 & -2 & -4 \\ -2 & 2 & 2 \\ -4 & 2 & 5 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}.
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Problem 2.1.6

Let A∈Rn×nA \in \mathbb {R}^{n \times n} and x→,b→∈Rn\overrightarrow {x}, \overrightarrow {b} \in \mathbb {R}^{n}. Consider the linear equation Ax→=b→A\overrightarrow {x} = \overrightarrow {b}. Show that it can be written as x→=Tx→\overrightarrow {x} = T\overrightarrow {x}, i.e. find Tx→T\overrightarrow {x}.

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Problem 2.1.7

If the system of linear equations Ax→=b→A\overrightarrow {x} = \overrightarrow {b} admits no solution we call the equations inconsistent. If there is a solution, the equations are called consistent. Let Ax→=b→A\overrightarrow {x} = \overrightarrow {b} be a system of mm linear equations in nn unknowns and suppose that the rank of AA is mm. Show that in this case Ax→=b→A\overrightarrow {x} = \overrightarrow {b} is consistent.

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Problem 2.1.8

Find all solutions of the linear system of three equations and four unknowns

x1+2x2−4x3+x4=3,2x1−3x2+x3+5x4=−4,7x1−10x3+13x4=0. x_{1}+2 x_{2}-4 x_{3}+x_{4} = 3, \qquad 2 x_{1}-3 x_{2}+x_{3}+5 x_{4} = -4, \qquad 7 x_{1}-10 x_{3}+13 x_{4} = 0.
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Problem 2.1.9
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(i)

Solve the linear equation

(x1x2x3)(123123123)=(x1x2x3). \begin{pmatrix} x_{1} & x_{2} & x_{3} \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} x_{1} & x_{2} & x_{3} \end{pmatrix}.
(ii)

Find x,y∈Rx, y \in \mathbb {R} such that

(2−3−12)=(3001)(2/3xy2). \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2/3 & x \\ y & 2 \end{pmatrix}.
Problem 2.1.10

Show that the curve fitting problem

j01234tj−1.0−0.50.00.51.0yj1.00.50.00.52.0 \begin{array}{c|ccccc} j & 0 & 1 & 2 & 3 & 4 \\ \hline t_{j} & -1.0 & -0.5 & 0.0 & 0.5 & 1.0 \\ y_{j} & 1.0 & 0.5 & 0.0 & 0.5 & 2.0 \end{array}

by a quadratic polynomial of the form p(t)=a2t2+a1t+a0p(t) = a_{2} t^{2}+a_{1} t+a_{0} leads to an overdetermined linear system.

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Problem 2.1.11

Consider the overdetermined linear system Ax→=b→A\overrightarrow {x} = \overrightarrow {b}. Find an x→^\hat{\overrightarrow {x}} such that

∥Ax→^−b→∥2=min⁡x→∥Ax→−b→∥2≡min⁡x→∥r→(x→)∥2 \left\| A\hat{\overrightarrow {x}}-\overrightarrow {b}\right\| _{2} = \min _{\overrightarrow {x}} \left\| A\overrightarrow {x}-\overrightarrow {b}\right\| _{2} \equiv \min _{\overrightarrow {x}} \left\| \overrightarrow {r}(\overrightarrow {x})\right\| _{2}

with the residual vector r→(x→):=b→−Ax→\overrightarrow {r}(\overrightarrow {x}) := \overrightarrow {b}-A\overrightarrow {x} and ∥⋅∥2\left\| \cdot \right\|_{2} denotes the Euclidean norm.

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Problem 2.1.12

Consider the overdetermined linear system Ax→=b→A\overrightarrow {x} = \overrightarrow {b} with

A=(111213141516171819110),x→=(x1x2),b→=(444458478493506516523531543571). A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 1 & 7 \\ 1 & 8 \\ 1 & 9 \\ 1 & 10 \end{pmatrix}, \qquad \overrightarrow {x} = \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}, \qquad \overrightarrow {b} = \begin{pmatrix} 444 \\ 458 \\ 478 \\ 493 \\ 506 \\ 516 \\ 523 \\ 531 \\ 543 \\ 571 \end{pmatrix}.

Solve this linear system in the least squares sense (see the previous problem) by the normal equations method.

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Problem 2.1.13

An underdetermined linear system is either inconsistent or has infinitely many solutions. Consider the underdetermined linear system Hx→=y→H\overrightarrow {x} = \overrightarrow {y}, where HH is an n×mn \times m matrix with m>nm>n and

x→=(x1x2⋮xm),y→=(y1y2⋮yn). \overrightarrow {x} = \begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{pmatrix}, \qquad \overrightarrow {y} = \begin{pmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{pmatrix}.

Assume that Hx→=y→H\overrightarrow {x} = \overrightarrow {y} has infinitely many solutions. Let QQ be the n×mn \times m matrix

Q=(10…00…001…00…0⋮⋮⋱⋮⋮⋮00…10…0). Q = \begin{pmatrix} 1 & 0 & \ldots & 0 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & & \vdots \\ 0 & 0 & \ldots & 1 & 0 & \ldots & 0 \end{pmatrix}.

We define x→^:=Qx→\hat{\overrightarrow {x}} := Q\overrightarrow {x}. Find

min⁡x→∥Qx→−y→∥22 \min _{\overrightarrow {x}} \left\| Q\overrightarrow {x}-\overrightarrow {y}\right\| _{2}^{2}

subject to the constraint ∥Hx→−y→∥22=0\left\| H\overrightarrow {x}-\overrightarrow {y}\right\|_{2}^{2} = 0. Assume that (λHTH+QTQ)−1(\lambda H^{T} H+Q^{T} Q)^{-1} exists for all λ>0\lambda >0. Apply the Lagrange multiplier method.

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Problem 2.1.14

Show that solving the system of nonlinear equations with the unknowns x1,x2,x3,x4x_{1}, x_{2}, x_{3}, x_{4}

(x1−1)2+(x2−2)2+x32=a2(x4−b1)2(x1−2)2+x22+(x3−2)2=a2(x4−b2)2(x1−1)2+(x2−1)2+(x3−1)2=a2(x4−b3)2(x1−2)2+(x2−1)2+x32=a2(x4−b4)2 \begin{aligned} (x_{1}-1)^{2}+(x_{2}-2)^{2}+x_{3}^{2} & = a^{2}(x_{4}-b_{1})^{2} \\ (x_{1}-2)^{2}+x_{2}^{2}+(x_{3}-2)^{2} & = a^{2}(x_{4}-b_{2})^{2} \\ (x_{1}-1)^{2}+(x_{2}-1)^{2}+(x_{3}-1)^{2} & = a^{2}(x_{4}-b_{3})^{2} \\ (x_{1}-2)^{2}+(x_{2}-1)^{2}+x_{3}^{2} & = a^{2}(x_{4}-b_{4})^{2} \end{aligned}

leads to a linear underdetermined system. Solve this system with respect to x1x_{1}, x2x_{2} and x3x_{3}.

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Problem 2.1.15

Let AA be an m×nm \times n matrix over R\mathbb {R}. We define

NA:={x→∈Rn:Ax→=0→}. N_{A} := \left\{ \overrightarrow {x} \in \mathbb {R}^{n} : A\overrightarrow {x} = \overrightarrow {0} \right\} .

NAN_{A} is called the kernel of AA and ν(A):=dim⁡(NA)\nu (A) := \dim (N_{A}) is called the nullity of AA. If NAN_{A} only contains the zero vector, then ν(A)=0\nu (A) = 0.

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(i)

Let

A=(12−12−13). A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \end{pmatrix}.

Find NAN_{A} and ν(A)\nu (A).

(ii)

Let

A=(2−134−26−63−9). A = \begin{pmatrix} 2 & -1 & 3 \\ 4 & -2 & 6 \\ -6 & 3 & -9 \end{pmatrix}.

Find NAN_{A} and ν(A)\nu (A).

Problem 2.1.16
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(i)

Let x1,x2,x3∈Zx_{1}, x_{2}, x_{3} \in \mathbb {Z}. Find all solutions of the system of linear equations

7x1+5x2−5x3=8,17x1+10x2−15x3=−42. 7 x_{1}+5 x_{2}-5 x_{3} = 8, \qquad 17 x_{1}+10 x_{2}-15 x_{3} = -42.
(ii)

Find all positive solutions.

Problem 2.1.17

Consider the inhomogeneous linear integral equation

∫01(α1(x)β1(y)+α2(x)β2(y))φ(y) dy+f(x)=φ(x)(1) \int _{0}^{1} (\alpha _{1}(x) \beta _{1}(y)+\alpha _{2}(x) \beta _{2}(y)) \varphi (y) \, dy+f(x) = \varphi (x) \tag {1}

for the unknown function φ\varphi, f(x)=xf(x) = x and

α1(x)=x,α2(x)=x,β1(y)=y,β2(y)=y. \alpha _{1}(x) = x, \qquad \alpha _{2}(x) = \sqrt{x}, \qquad \beta _{1}(y) = y, \qquad \beta _{2}(y) = \sqrt{y}.

Thus α1\alpha_{1} and α2\alpha_{2} are continuous in [0,1][0,1] and likewise for β1\beta_{1} and β2\beta_{2}. We define

B1:=∫01β1(y)φ(y) dy,B2:=∫01β2(y)φ(y) dy B_{1} := \int _{0}^{1} \beta _{1}(y) \varphi (y) \, dy, \qquad B_{2} := \int _{0}^{1} \beta _{2}(y) \varphi (y) \, dy

and

aμν:=∫01βμ(y)αν(y) dy,bμ:=∫01βμ(y)f(y) dy a_{\mu \nu } := \int _{0}^{1} \beta _{\mu }(y) \alpha _{\nu }(y) \, dy, \qquad b_{\mu } := \int _{0}^{1} \beta _{\mu }(y) f(y) \, dy

where μ,ν=1,2\mu , \nu = 1, 2. Show that the integral equation can be cast into a system of linear equations for B1B_{1} and B2B_{2}. Solve this system of linear equations and thus find a solution of the integral equation.

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Problem 2.1.18

Consider the area-preserving map of the two-dimensional torus (modulo 1)

(x1′x2′)=A(x1x2),A=(41514) \begin{pmatrix} x_{1}' \\ x_{2}' \end{pmatrix} = A \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}, \qquad A = \begin{pmatrix} 4 & 15 \\ 1 & 4 \end{pmatrix}

where det⁡(A)=1\operatorname {det}\left(A\right) = 1 (area-preserving). Consider a rational point on the torus

(x1x2)=(n1/pn2/p) \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} n_{1}/p \\ n_{2}/p \end{pmatrix}

where pp is a prime number (except 2, 3, 5) and n1n_{1}, n2n_{2} are integers between 0 and p−1p-1. One finds that the orbit has the following property. It is periodic and its period TT depends on pp alone. Consider p=7p = 7, n1=2n_{1} = 2, n2=3n_{2} = 3. Find the orbit and the period TT.

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Problem 2.1.19

Let AA be an m×nm \times n matrix over R\mathbb {R} and c→\overrightarrow {c} be an nn-vector in Rn\mathbb {R}^{n}.

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(i)

Let

A=(101010101). A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}.

Find out whether system (1) or system (2) has a solution.

(ii)

Let

A=(211121112). A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}.

Find out whether system (1) or system (2) has a solution.

Remark.
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Gordan's theorem tells us the following. Then exactly one of the following systems has a solution: System 1: Ax→<0→A\overrightarrow {x}<\overrightarrow {0} for some x→∈Rn\overrightarrow {x} \in \mathbb {R}^{n}. System 2: ATp→=0→A^{T} \overrightarrow {p} = \overrightarrow {0} and p→≥0→\overrightarrow {p} \geq \overrightarrow {0} for some p→∈Rm\overrightarrow {p} \in \mathbb {R}^{m}.

Problem 2.1.20

Let

A=(101010101),c→=(111). A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \qquad \overrightarrow {c} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.

Find out whether system (1) or system (2) has a solution.

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Remark.
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Farkas' theorem tells us the following. Let AA be an m×nm \times n matrix over R\mathbb {R} and c→\overrightarrow {c} be an nn-vector in Rn\mathbb {R}^{n}. Then exactly one of the following systems has a solution: System 1: Ax→≤0→A\overrightarrow {x} \leq \overrightarrow {0} and c→Tx→>0\overrightarrow {c}^{T} \overrightarrow {x}>0 for some x→∈Rn\overrightarrow {x} \in \mathbb {R}^{n}. System 2: ATy→=c→A^{T} \overrightarrow {y} = \overrightarrow {c} and y→≥0→\overrightarrow {y} \geq \overrightarrow {0} for some y→∈Rm\overrightarrow {y} \in \mathbb {R}^{m}.

Problem 2.1.21

Let AA be an n×nn \times n matrix. Consider the linear equation Ax→=0→A\overrightarrow {x} = \overrightarrow {0}. If the matrix AA has rank rr, then there are n−rn-r linearly independent solutions of Ax→=0→A\overrightarrow {x} = \overrightarrow {0}. Let n=3n = 3 and

A=(011001000). A = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.

Find the rank of AA and the linearly independent solutions.

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Problem 2.1.22

Consider the curve described by the equation

2x2+4xy−y2+4x−2y+5=0(1) 2 x^{2}+4 x y-y^{2}+4 x-2 y+5 = 0 \tag {1}

relative to the natural basis (standard basis e→1=(10)T\overrightarrow {e}_{1} = \begin{pmatrix} 1 & 0 \end{pmatrix}^{T}, e→2=(01)T\overrightarrow {e}_{2} = \begin{pmatrix} 0 & 1 \end{pmatrix}^{T}).

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(i)

Write the equation in matrix form.

(ii)

Find an orthogonal change of basis so that the equation relative to the new basis has no crossterms, i.e. no x′y′x'y' term. This change of coordinate system does not change the origin.

Problem 2.1.23

Consider the 2×22 \times 2 matrix

(b−aab) \begin{pmatrix} b & -a \\ a & b \end{pmatrix}

with a,b∈Ra, b \in \mathbb {R} and positive determinant, i.e. a2+b2>0a^{2}+b^{2}>0. Solve the equation

(b−aab)(x1y1)=(ba−ab)(x0y0) \begin{pmatrix} b & -a \\ a & b \end{pmatrix} \begin{pmatrix} x_{1} \\ y_{1} \end{pmatrix} = \begin{pmatrix} b & a \\ -a & b \end{pmatrix} \begin{pmatrix} x_{0} \\ y_{0} \end{pmatrix}

for the vector (x1y1)T\begin{pmatrix} x_{1} & y_{1} \end{pmatrix}^{T} with a given vector (x0y0)T\begin{pmatrix} x_{0} & y_{0} \end{pmatrix}^{T}.

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Problem 2.1.24

Let VV be a vector space over a field R\mathbb {R}. Let WW be a subspace of VV. We define an equivalence relation ∼\sim on VV by stating that v1∼v2v_{1} \sim v_{2} if v1−v2∈Wv_{1}-v_{2} \in W. The quotient space V/WV/W is the set of equivalence classes [v][v] where v1−v2∈Wv_{1}-v_{2} \in W. Thus we can say that v1v_{1} is equivalent to v2v_{2} modulo WW if v1=v2+wv_{1} = v_{2}+w for some w∈Ww \in W. Let

V=R2={(x1x2):x1,x2∈R} V = \mathbb {R}^{2} = \left\{ \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} : x_{1}, x_{2} \in \mathbb {R} \right\}

and the subspace

W={(x10):x1∈R}. W = \left\{ \begin{pmatrix} x_{1} \\ 0 \end{pmatrix} : x_{1} \in \mathbb {R} \right\} .
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(i)

Is

(30)∼(10),(41)∼(−31),(30)∼(41)  ? \begin{pmatrix} 3 \\ 0 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad \begin{pmatrix} 4 \\ 1 \end{pmatrix} \sim \begin{pmatrix} -3 \\ 1 \end{pmatrix}, \qquad \begin{pmatrix} 3 \\ 0 \end{pmatrix} \sim \begin{pmatrix} 4 \\ 1 \end{pmatrix} \; ?
(ii)

Give the quotient space V/WV/W.

Problem 2.1.25

Suppose that VV is a vector space over a field R\mathbb {R} and U⊂VU \subset V is a subspace. We define an equivalence relation ∼\sim on VV by x∼yx \sim y iff x−y∈Ux-y \in U. Let V/U=V/∼V/U = V/\sim. Define addition and scalar multiplication on V/UV/U by [x]+[y]=[x]+[y][x]+[y] = [x]+[y], c[x]=[cx]c[x] = [cx], where c∈Rc \in \mathbb {R} and

[x]={y∈V:y∼x}. [x] = \left\{ y \in V : y \sim x \right\} .
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(i)

Show that these operations do not depend on which representative xx we choose.

(ii)

Let V=C2V = \mathbb {C}^{2} and the subspace U={(x1,x2):x1=2x2}U = \left\{ (x_{1}, x_{2}) : x_{1} = 2 x_{2} \right\}. Find V/UV/U.

Problem 2.1.26

Let b>ab>a. Consider the system of linear equations

(111…1x0x1x2⋯xnx02x12x22…xn2⋮⋮⋮⋱⋮x0nx1nx2n…xnn)(w0w1w2⋮wn)=(b−a(b2−a2)/2(b3−a3)/3⋮(bn+1−an+1)/(n+1)). \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 \\ x_{0} & x_{1} & x_{2} & \cdots & x_{n} \\ x_{0}^{2} & x_{1}^{2} & x_{2}^{2} & \ldots & x_{n}^{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{0}^{n} & x_{1}^{n} & x_{2}^{n} & \ldots & x_{n}^{n} \end{pmatrix} \begin{pmatrix} w_{0} \\ w_{1} \\ w_{2} \\ \vdots \\ w_{n} \end{pmatrix} = \begin{pmatrix} b-a \\ (b^{2}-a^{2})/2 \\ (b^{3}-a^{3})/3 \\ \vdots \\ (b^{n+1}-a^{n+1})/(n+1) \end{pmatrix}.

Let n=2n = 2, a=0a = 0, b=1b = 1, x0=0x_{0} = 0, x1=1/2x_{1} = 1/2, x2=1x_{2} = 1. Find w0,w1,w2w_{0}, w_{1}, w_{2}.

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Problem 2.1.27

Let Y,X,A,B,C,EY, X, A, B, C, E n×nn \times n matrices over R\mathbb {R}. Consider the system of matrix equations Y+CE+DX=0nY+C E+D X = 0_{n}, AE+BX=0nA E+B X = 0_{n}. Assume that AA has an inverse. Eliminate the matrix EE and solve the system for YY.

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Problem 2.1.28

For the three-body problem the following linear transformation plays a role

X(x1,x2,x3)=13(x1+x2+x3)x(x1,x2,x3)=12(x1−x2)y(x1,x2,x3)=16(x1+x2−2x3). \begin{aligned} X(x_{1}, x_{2}, x_{3}) & = \frac{1}{3}(x_{1}+x_{2}+x_{3}) \\ x(x_{1}, x_{2}, x_{3}) & = \frac{1}{\sqrt{2}}(x_{1}-x_{2}) \\ y(x_{1}, x_{2}, x_{3}) & = \frac{1}{\sqrt{6}}(x_{1}+x_{2}-2 x_{3}). \end{aligned}
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(i)

Find the inverse transformation.

(ii)

Introduce polar coordinates x(r,ϕ)=rsin⁡(ϕ)x(r,\phi ) = r\sin (\phi ), y(r,ϕ)=rcos⁡(ϕ)y(r,\phi ) = r\cos (\phi ) and

r2=13((x1−x2)2+(x2−x3)2+(x3−x1)2). r^{2} = \frac{1}{3} \left( (x_{1}-x_{2})^{2}+(x_{2}-x_{3})^{2}+(x_{3}-x_{1})^{2} \right).

Express (x1−x2)(x_{1}-x_{2}), (x2−x3)(x_{2}-x_{3}), (x3−x1)(x_{3}-x_{1}) using these three coordinates.

Problem 2.1.29

Let α∈[0,2π)\alpha \in [0, 2\pi ). Find all solutions of the linear equation

(cos⁡(α)sin⁡(α)sin⁡(α)cos⁡(α))(x1x2)=(11). \begin{pmatrix} \cos (\alpha ) & \sin (\alpha ) \\ \sin (\alpha ) & \cos (\alpha ) \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

Thus x1x_{1} and x2x_{2} depend on α\alpha.

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Problem 2.1.30

Consider the partial differential equation (Laplace equation)

∂2u∂x2+∂2u∂y2=0on[0,1]×[0,1] \frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = 0 \quad \text{on} \quad [0,1] \times [0,1]

with the boundary conditions

u(x,0)=1,u(x,1)=2,u(0,y)=1,u(1,y)=2. u(x,0) = 1, \qquad u(x,1) = 2, \qquad u(0,y) = 1, \qquad u(1,y) = 2.

Apply the central difference scheme

(∂2u∂x2)j,k≈uj−1,k−2uj,k+uj+1,k(Δx)2,(∂2u∂y2)j,k≈uj,k−1−2uj,k+uj,k+1(Δy)2 \left( \frac{\partial ^{2} u}{\partial x^{2}} \right)_{j,k} \approx \frac{u_{j-1,k}-2 u_{j,k}+u_{j+1,k}}{(\Delta x)^{2}}, \qquad \left( \frac{\partial ^{2} u}{\partial y^{2}} \right)_{j,k} \approx \frac{u_{j,k-1}-2 u_{j,k}+u_{j,k+1}}{(\Delta y)^{2}}

and then solve the linear equation. Consider the cases Δx=Δy=1/3\Delta x = \Delta y = 1/3 and Δx=Δy=1/4\Delta x = \Delta y = 1/4.

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Problem 2.1.31
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(i)

The equation of a line in the Euclidean space R2\mathbb {R}^{2} passing through the points (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}) is given by

(y−y1)(x2−x1)=(y2−y1)(x−x1). (y-y_{1})(x_{2}-x_{1}) = (y_{2}-y_{1})(x-x_{1}).

Apply this equation to the points in R2\mathbb {R}^{2} given by (x1,y1)=(1,1/2)(x_{1}, y_{1}) = (1, 1/2), (x2,y2)=(1/2,1)(x_{2}, y_{2}) = (1/2, 1). Consider the unit square with the corner points (0,0)(0,0), (0,1)(0,1), (1,0)(1,0), (1,1)(1,1) and the map

(0,0)→0,(0,1)→0,(1,0)→0,(1,1)→1. (0,0) \to 0, \quad (0,1) \to 0, \quad (1,0) \to 0, \quad (1,1) \to 1.

We can consider this as a 2 input AND-gate. Show that the line constructed above classifies this map.

(ii)

The equation of a plane in R3\mathbb {R}^{3} passing through the points (x1,y1,z1)(x_{1}, y_{1}, z_{1}), (x2,y2,z2)(x_{2}, y_{2}, z_{2}), (x3,y3,z3)(x_{3}, y_{3}, z_{3}) in R3\mathbb {R}^{3} is given by

det⁡((x−x1y−y1z−z1x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1))=0. \operatorname {det}\left( \begin{pmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{pmatrix} \right) = 0.

Apply this equations to the points (1,1,1/2)(1, 1, 1/2), (1,1/2,1)(1, 1/2, 1), (1/2,1,1)(1/2, 1, 1). Consider the unit cube in R3\mathbb {R}^{3} with the corner points (vertices)

(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1) (0,0,0), \quad (0,0,1), \quad (0,1,0), \quad (0,1,1), \quad (1,0,0), \quad (1,0,1), \quad (1,1,0), \quad (1,1,1)

and the map where all corner points are mapped to 0 except for (1,1,1)(1,1,1) which is mapped to 1. We can consider this as a 3 input AND-gate. Show that the plane constructed in (i) separates these solutions.

Problem 2.1.32

Find the system of linear equations for aa and bb given by

x+15(x+3)(x−1)=ax+3+bx−1. \frac{x+15}{(x+3)(x-1)} = \frac{a}{x+3}+\frac{b}{x-1}.

Solve the system of linear equations.

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Problem 2.1.33

Let k=1,2,3,4k = 1, 2, 3, 4 and x0=1x_{0} = 1, x5=0x_{5} = 0. Solve xk−1−2xk+xk+1=0x_{k-1}-2 x_{k}+x_{k+1} = 0.

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Problem 2.1.34

Consider the polynomial p(x)=a+bx+cx2p(x) = a+bx+cx^{2}. Find a,b,ca, b, c from the conditions p(0)=0p(0)=0, p(1)=1p(1)=1, p(2)=0p(2)=0.

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Problem 2.1.35

Let A∈Cn×mA \in \mathbb {C}^{n \times m} with n≥mn \geq m and rk⁡(A)=m\operatorname {rk}\left(A\right) = m.

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(i)

Show that the m×mm \times m matrix A∗AA^{*} A is invertible.

(ii)

We set Π:=A(A∗A)−1A\Pi := A(A^{*} A)^{-1} A. Show that Π\Pi is a projection matrix, i.e. Π2=Π\Pi^{2} = \Pi and Π=Π∗\Pi = \Pi^{*}.

Problem 2.1.36

Let a,b,c∈Ra, b, c \in \mathbb {R} and abc≠0abc \neq 0. Find the solution of the system of linear equations

(0cbc0aba0)(cos⁡(α)cos⁡(β)cos⁡(γ))=(abc). \begin{pmatrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \end{pmatrix} \begin{pmatrix} \cos (\alpha ) \\ \cos (\beta ) \\ \cos (\gamma ) \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.
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Problem 2.1.37

Let AA be an invertible n×nn \times n matrix over R\mathbb {R}. Consider the system of linear equation Ax→=b→A\overrightarrow {x} = \overrightarrow {b} or

∑j=1naijxj=bi,i=1,…,n. \sum _{j=1}^{n} a_{ij} x_{j} = b_{i}, \qquad i = 1, \ldots , n.

Let A=C−RA = C-R. This is called a splitting of the matrix AA and RR is the defect matrix of the splitting. Consider the iteration

Cx→(k+1)=Rx→(k)+b→,k=0,1,2,…. C \overrightarrow {x}^{(k+1)} = R \overrightarrow {x}^{(k)}+\overrightarrow {b}, \qquad k = 0, 1, 2, \ldots .

Let

A=(4−10−14−10−24),C=(400040004),b→=(322),x→(0)=(000). A = \begin{pmatrix} 4 & -1 & 0 \\ -1 & 4 & -1 \\ 0 & -2 & 4 \end{pmatrix}, \qquad C = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}, \qquad \overrightarrow {b} = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}, \qquad \overrightarrow {x}^{(0)} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.

The iteration converges if ρ(C−1R)<1\rho (C^{-1} R)<1, where ρ(C−1R)\rho (C^{-1} R) denotes the spectral radius of C−1RC^{-1} R. Show that ρ(C−1R)<1\rho (C^{-1} R)<1. Perform the iteration.

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Problem 2.1.38

Let AA be an n×nn \times n matrix over R\mathbb {R} and let b→∈Rn\overrightarrow {b} \in \mathbb {R}^{n}. Consider the linear equation Ax→=b→A\overrightarrow {x} = \overrightarrow {b}. Assume that ajj≠0a_{jj} \neq 0 for j=1,…,nj = 1, \ldots , n. We define the diagonal matrix D=diag⁡(ajj)D = \operatorname {diag}(a_{jj}). Then the linear equation Ax→=b→A\overrightarrow {x} = \overrightarrow {b} can be written as

x→=Bx→+c→ \overrightarrow {x} = B\overrightarrow {x}+\overrightarrow {c}

with B:=−D−1(A−D)B := -D^{-1}(A-D), c→:=D−1b→\overrightarrow {c} := D^{-1} \overrightarrow {b}. The Jacobi method for the solution of the linear equation Ax→=b→A\overrightarrow {x} = \overrightarrow {b} is given by

x→(k+1)=Bx→(k)+c→,k=0,1,… \overrightarrow {x}^{(k+1)} = B \overrightarrow {x}^{(k)}+\overrightarrow {c}, \qquad k = 0, 1, \ldots

where x→(0)\overrightarrow {x}^{(0)} is any initial vector in Rn\mathbb {R}^{n}. The sequence converges if

ρ(B):=max⁡j=1,…,n∣λj(B)∣<1 \rho (B) := \max _{j=1,\ldots ,n} \left|\lambda _{j}(B)\right| < 1

where ρ(B)\rho (B) is the spectral radius of BB. Let

A=(210121012). A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}.
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(i)

Show that the Jacobi method can applied for this matrix.

(ii)

Find the solution of the linear equation with b→=(111)T\overrightarrow {b} = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix}^{T}.

Problem 2.1.39

Kirchhoff’s current law states that the algebraic sum of all the currents flowing into a junction is 0. Kirchhoff’s voltage law states that the algebraic sum of all the voltages around a closed circuit is 0. Use Kirchhoff's laws and Ohm’s law (V=RIV = RI) to setting up the system of linear equations for the circuit depicted in the figure.

Circuit for Problem 2.1.39Circuit for Problem 2.1.39

Given the voltages VAV_{A}, VBV_{B} and the resistors R1R_{1}, R2R_{2}, R3R_{3}. Find I1I_{1}, I2I_{2}, I3I_{3}.

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Problem 2.1.40

Write a C++ program that implements Gauss elimination to solve linear equations. Apply it to the system

(1118422793)(x1x2x3)=(1514). \begin{pmatrix} 1 & 1 & 1 \\ 8 & 4 & 2 \\ 27 & 9 & 3 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ 14 \end{pmatrix}.
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Problem 2.2.1

Let n→\overrightarrow {n} and p→\overrightarrow {p} be vectors in Rn\mathbb {R}^{n} with n→≠0→\overrightarrow {n} \neq \overrightarrow {0}. The set of all vectors x→\overrightarrow {x} in Rn\mathbb {R}^{n} which satisfy the equation n→⋅(x→−p→)=0\overrightarrow {n} \cdot (\overrightarrow {x}-\overrightarrow {p}) = 0 is called a hyperplane through the point p→∈R\overrightarrow {p} \in \mathbb {R}. We call n→\overrightarrow {n} a normal vector for the hyperplane and call n→⋅(x→−p→)=0\overrightarrow {n} \cdot (\overrightarrow {x}-\overrightarrow {p}) = 0 a normal equation for the hyperplane. Find n→\overrightarrow {n} and p→\overrightarrow {p} in R4\mathbb {R}^{4} such that we obtain the hyperplane given by

x1+x2+x3+x4=72. x_{1}+x_{2}+x_{3}+x_{4} = \frac{7}{2}.

Any hyperplane of the Euclidean space Rn\mathbb {R}^{n} has exactly two unit normal vectors.

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Problem 2.2.2

The hyperplanes x1+x2+x3+x4=1x_{1}+x_{2}+x_{3}+x_{4} = 1, x1+x2+x3+x4=2x_{1}+x_{2}+x_{3}+x_{4} = 2 in R4\mathbb {R}^{4} do not intersect. Find the shortest distance.

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Problem 2.2.3
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(i)

Find the solutions of the equation

(1234)(x11x12x21x22)=(0000). \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}.
(ii)

Find the solutions of the equation

(1234)(x11x12x21x22)=(1001). \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.
Problem 2.2.4

Let AA be a given 3×33 \times 3 matrix over R\mathbb {R} with det⁡(A)≠0\operatorname {det}\left(A\right) \neq 0. Is the transformation

x1′(x1,x2)=a11x1+a12x2+a13a31x1+a32x2+a33,x2′(x1,x2)=a21x1+a22x2+a23a31x1+a32x2+a33 x_{1}'(x_{1}, x_{2}) = \frac{a_{11} x_{1}+a_{12} x_{2}+a_{13}}{a_{31} x_{1}+a_{32} x_{2}+a_{33}}, \qquad x_{2}'(x_{1}, x_{2}) = \frac{a_{21} x_{1}+a_{22} x_{2}+a_{23}}{a_{31} x_{1}+a_{32} x_{2}+a_{33}}

invertible? If so find the inverse.

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Problem 2.2.5
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(i)

Consider the linear equation written in matrix form

(101011102)(x1x2x3)=(121). \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}.

The determinant of the 3×33 \times 3 matrix is nonzero (+1+1) and the inverse matrix is

(20−111−1−101). \begin{pmatrix} 2 & 0 & -1 \\ 1 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}.

Apply two different methods (Gauss elimination and the Leverrier's method) to find the solution. Compare the two methods and discuss.

(ii)

Apply the Gauss-Seidel method to solve the linear system

(4−1−10−140−1−104−10−1−14)(x1x2x3x4)=(1000) \begin{pmatrix} 4 & -1 & -1 & 0 \\ -1 & 4 & 0 & -1 \\ -1 & 0 & 4 & -1 \\ 0 & -1 & -1 & 4 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}

and show that the solution is given by

x1=724,x2=112,x3=112,x4=124. x_{1} = \frac{7}{24}, \qquad x_{2} = \frac{1}{12}, \qquad x_{3} = \frac{1}{12}, \qquad x_{4} = \frac{1}{24}.
(iii)

Show that the solution of the system of linear equations

x1+x2+x3=0,x1+2x2+x3=1,2x1+x2+x3=2 x_{1}+x_{2}+x_{3} = 0, \qquad x_{1}+2 x_{2}+x_{3} = 1, \qquad 2 x_{1}+x_{2}+x_{3} = 2

is given by x1=2x_{1} = 2, x2=1x_{2} = 1, x3=−3x_{3} = -3.

Problem 2.2.6
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(i)

Given the 2×22 \times 2 matrix AA. Find all 2×22 \times 2 matrices XX such that AX=XAAX = XA. From AX=XAAX = XA we obtain

(a11x11+a12x21a11x12+a12x22a21x11+a22x21a21x12+a22x22)=(a11x11+a21x12a12x11+a22x12a11x21+a21x22a12x21+a22x22). \begin{pmatrix} a_{11} x_{11}+a_{12} x_{21} & a_{11} x_{12}+a_{12} x_{22} \\ a_{21} x_{11}+a_{22} x_{21} & a_{21} x_{12}+a_{22} x_{22} \end{pmatrix} = \begin{pmatrix} a_{11} x_{11}+a_{21} x_{12} & a_{12} x_{11}+a_{22} x_{12} \\ a_{11} x_{21}+a_{21} x_{22} & a_{12} x_{21}+a_{22} x_{22} \end{pmatrix}.

Thus we obtain the four equations

a12x21=a21x12,a11x12+a12x12=a12x11+a22x12 a_{12} x_{21} = a_{21} x_{12}, \qquad a_{11} x_{12}+a_{12} x_{12} = a_{12} x_{11}+a_{22} x_{12} a21x12=a12x21,a21x11+a22x21=a11x21+a21x22. a_{21} x_{12} = a_{12} x_{21}, \qquad a_{21} x_{11}+a_{22} x_{21} = a_{11} x_{21}+a_{21} x_{22}.
(ii)

Show that the solution of the equation

(0110)(x11x12x21x22)=(x11x12x21x22)(0110) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

is given by x11=x22x_{11} = x_{22}, x12=x21x_{12} = x_{21}.

Problem 2.2.7

Let α,b1,b2∈R\alpha , b_{1}, b_{2} \in \mathbb {R}. Solve the system of linear equations

(cosh⁡(α)sinh⁡(α)sinh⁡(α)cosh⁡(α))(x1x2)=(b1b2). \begin{pmatrix} \cosh (\alpha ) & \sinh (\alpha ) \\ \sinh (\alpha ) & \cosh (\alpha ) \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} = \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix}.
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