Question

1. Consider the matrix and vectors A=(: -5] -- [].x = [1] a. Show that the vectors v1 and v2 are eigenvectors of A and find their associated eigenvalues. Evaluate (Sage) D. Express the vector x = as a linear combination of vi and v2. c. Use this expression to compute Ax, APx, and A 'xas a linear combination of eigenvectors.

Answer 1

Solution:

$\small A =\begin{bmatrix} 8 &-10 \\ 5 &-7 \end{bmatrix} , v_{1} = \begin{bmatrix} 2\\ 1 \end{bmatrix}, v_{2} = \begin{bmatrix} 1\\ 1 \end{bmatrix}$

a)

A- 1 = 0

18- | 5 -10 -7-X

-(7 + 1)(8 - x)-(-50) = 0

—(56 – 7X+8X-X)-(-50) = 0

0=9-P-

12 - 3x + 2x -6=0

X(1-3) + 2(x-3) = 0

(1+2)(1-3) = 0

= -2,3

i) $\small \lambda_{1} =3$

$\small \begin{bmatrix} 5& -10\\ 5&-10 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$

→ 5.11 - 10.12 = 0

= I1 = 2.12

Let

$\small x_{2}= s$

I'1 = 2s

$\small \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} 2s\\ s \end{bmatrix}=\left \{ s\begin{bmatrix} 2\\ 1 \end{bmatrix} \right \}$

$\small v_{1}=\begin{bmatrix} 2\\ 1 \end{bmatrix}$

ii)

2 = -2

$\small \begin{bmatrix} 10& -10\\ 5&-5 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$

= 1021 - 10.x2 = 0

→ I1 = I2 = 7

$\small \begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} = \begin{bmatrix} t\\ t \end{bmatrix}=\left \{ t\begin{bmatrix} 1\\ 1 \end{bmatrix} \right \}$

$\small v_{2} =\begin{bmatrix} 1\\ 1 \end{bmatrix}$

$\small \therefore v_{1}=\begin{bmatrix} 2\\ 1 \end{bmatrix}$ and $\small v_{2} =\begin{bmatrix} 1\\ 1 \end{bmatrix}$ are eigen vectors of A and their corresponding eigen values are 3 and -2 respectively.

b)

$\small x=\begin{bmatrix} -4\\ -1 \end{bmatrix}$

Just put,  , then solve for a and b

1 = avi + bu

2a + b = -4

a + b = -1

→ a= -3

b=2

I= -301 +202

$\small \therefore$ x is a linear combination of $\small v_{1}$ and $\small v_{2}$ .

c)

i)

Ar = Co AS

Az = avi + buz

2a + b = -22

a + b = -13

6- = =

b=-4

Ar = -901 - 402

$\small \therefore$ Ax is a linear combination of $\small v_{1}$ and $\small v_{2}$ .

ii)

A2 = To -10

$\small A^{2}x= \begin{bmatrix} 14 &-10 \\ 5&1 \end{bmatrix}\begin{bmatrix} -4\\ -1 \end{bmatrix} = \begin{bmatrix} -46\\ -21 \end{bmatrix}$

AP1 = avi + buz

2a+b = -46

a +b =-21

→ a= -25

b=4

A x = -2501 + 402

$\small \therefore$$\small A^{2}x$ is a linear combination of  $\small v_{1}$ and  $\small v_{2}$.

iii)

det(A) = -6

$\small A^{-1} = \frac{1}{(-6)}\begin{bmatrix} -7 &10 \\ -5& 8 \end{bmatrix}=\begin{bmatrix} 7/6 &-5/3 \\ 5/6&-4/3 \end{bmatrix}$

$\small A^{-1}x =\begin{bmatrix} 7/6 &-5/3 \\ 5/6&-4/3 \end{bmatrix}\begin{bmatrix} -4\\ -1 \end{bmatrix} = \begin{bmatrix} -3\\ -2 \end{bmatrix}$

$\small A^{-1}x = av_{1} + bv_{2}$

2a+b = -3

a+b = -2

$\small \Rightarrow a=-1$

$\small \Rightarrow b=-1$

A-?r= -01 - 02

$\small \therefore$  $\small A^{-1}x$ is a linear combination of  $\small v_{1}$ and  $\small v_{2}$.

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