Question

(1 point) Find the eigenvalues of the matrix A . -19 6 0 0 -36 11 0 0 A= The eigenvalues are λ| < λ2 < λ3 < λ4, where has an eigenvector 12 has an eigenvector has an eigenvector 4 has an eigenvector Note: you may want to use a graphing calculator to estimate the roots of the polynomial which defines the eigenvalues

Answer 1

Given matrix is : A =

Now the characteristic equation of A is : $\begin{vmatrix} -19-x & 6 & 0 & 0\\ -36 & 11-x & 0 & 0\\ 0 & 0 & 8-x & 7\\ 0 & 0 & 0 & 1-x \end{vmatrix}=0$

i.e., [(-19-x)*(11-x)-6*(-36)]*[(8-x)*(1-x)-7*0] = 0

i.e., (8-x)(1-x)[x²+8x-209+216] = 0

i.e., (8-x)(1-x)(x²+8x+7) = 0

i.e., (8-x)(1-x)(x+1)(x+7) = 0

i.e., x = -7, -1, 1, 8

Therefore, the eigenvalues of A are -7, -1, 1, 8.

For $\lambda_1$ = -7 : (A+7I)X = 0

i.e.,
12 6 0 0 一36 18 0 0 0 0 15 7 0 0 0 8 3

ab cd
= $\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

i.e., -12a+6b = 0

-36a+18b = 0

15c+7d = 0

8d = 0

i.e., b = 2a

c = 0

d = 0

Let us take a = k. Then, b = 2k.

Therefore, eigenvectors corresponding to eigenvalue -7 are $k\begin{bmatrix} 1\\ 2\\ 0\\ 0 \end{bmatrix}$ , where k is real number.

For $\lambda_2$ = -1 : (A+I)X = 0

i.e., $\begin{bmatrix} -18 & 6 & 0 & 0\\ -36 & 12 & 0 & 0\\ 0 & 0 & 9 & 7\\ 0 & 0 & 0 & 2 \end{bmatrix}$
ab cd
= $\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

i.e., -18a+6b = 0

-36a+12b = 0

9c+7d = 0

2d = 0

i.e., b = 3a

c = 0

d = 0

Let us take a = k. Then, b = 3k.

Therefore, eigenvectors corresponding to eigenvalue -1 are $k\begin{bmatrix} 1\\ 3\\ 0\\ 0 \end{bmatrix}$ , where k is real number.

For $\lambda_3$ = 1 : (A-I)X = 0

i.e., $\begin{bmatrix} -20 & 6 & 0 & 0\\ -36 & 10 & 0 & 0\\ 0 & 0 & 7 & 7\\ 0 & 0 & 0 & 0 \end{bmatrix}$
ab cd
= $\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

i.e., -20a+6b = 0

-36a+10b = 0

7c+7d = 0

i.e., a = 0

b = 0

d = -c

Let us take c = k. Then, d = -k.

Therefore, eigenvectors corresponding to eigenvalue 1 are $k\begin{bmatrix} 0\\ 0\\ 1\\ -1 \end{bmatrix}$ , where k is real number.

For $\lambda_4$ = 8 : (A-8I)X = 0

i.e., $\begin{bmatrix} -27 & 6 & 0 & 0\\ -36 & 3 & 0 & 0\\ 0 & 0 & 0 & 7\\ 0 & 0 & 0 & -7 \end{bmatrix}$
ab cd
= $\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$

i.e., -27a+6b = 0

-36a+3b = 0

7d = 0

-7d = 0

i.e., a = 0

b = 0

d = 0

Let us take c = k.

Therefore, eigenvectors corresponding to eigenvalue 8 are $k\begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix}$ , where k is real number.