How to find an eigenvector that corresponds to it.

Question

A=$\begin{pmatrix} -7 & -9 \\ 6 & 8 \end{pmatrix}$

(a) Is the vector v = $\left[ \begin{matrix} 1 \\ -1 \end{matrix} \right] $ an eigenvector of A? If so, what is the corresponding eigenvalue?

(b) Given that −1 is an eigenvalue of A, find an eigenvector that corresponds to it.

My work

I got the eigen values as $-10,11$

But, how to find part b).

I am getting the eigen vector as $\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] $

Answer 1

Since $A.\binom1{-1}=\binom2{-2}=2\binom1{-1}$, $v$ is an eigenvector and its eigenvalue is $2$. The other eigenvalue is $-1$. A corresponding eigenvector is $\binom3{-2}$.

Answer 2

You have for every eigenvector the relation $Ax = \lambda x$. This is the same as saying $(A-\lambda I)x = 0$ for different couples of $\lambda $ and $x$. So in your case, you simply want to find the vectors $x$ which satisfies the above relation for $ \lambda = -1$. That is, you want to solve $(A+I)x=0$.

Answer 3

Do the multiplication $$ A\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}2\\-2\end{bmatrix} $$ Is the result a scalar multiple of $\begin{bmatrix}1\\-1\end{bmatrix}$?

For the other eigenvalue, find the null space of $$ A+I=\begin{bmatrix} -6 & -9 \\ 6 & 9 \end{bmatrix} $$

Since the characteristic polynomial is $X^2-\operatorname{trace}(A)X+\det(A)=X^2-X-2$, it's difficult to know how you computed the eigenvalues as $-10$ and $11$.

Answer 4

See below to check where you got off the track and got $-10,11$.

Eigenvalues: $$\begin{vmatrix} -7-\lambda& -9 \\ 6 & 8-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-\lambda -2=0 \Rightarrow \lambda_1=-1, \lambda_2=2.$$ Eigenvector for $\lambda=-1$: $$\begin{pmatrix}-7+1 & -9\\ 6&8+1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=0\Rightarrow 6x_1+9x_2=0 \Rightarrow x_2=-\frac{2}{3}x_1 \Rightarrow \\ \begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}3\\-2\end{pmatrix},\begin{pmatrix}1\\-\frac 23\end{pmatrix},\begin{pmatrix} -\frac 32\\1\end{pmatrix},\cdots$$

Eigenvector for $\lambda=2$: $$\begin{pmatrix}-7-2 & -9\\ 6&8-2\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=0\Rightarrow 6x_1+6x_2=0 \Rightarrow x_2=-x_1 \Rightarrow \begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix}$$