linear system straight line solution

Question

For a matrix

$ A = \begin{bmatrix} -1 & 3 \\ 0 & -2 \end{bmatrix} $

I have obtained the Eigenvalues $\lambda = -2$ and $\lambda = -1$.

Now, after trying to find the straight line solution for $\lambda = -1$, I get:

$ \begin{bmatrix} 0 & 3 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

...

Now what does it mean? $ 3y = 0, -y = 0$, but what is the eigenvector ? ? ?

Can it be a vector (0,1) ?

Thanks.

Answer 1

$3y=0$ and $-y=0$ implies $y=0$. For a linear transformation $T$ representing the matrix, $\IM T = \{(0,0)\}$ for every $x$. So for every $x,y$ you get $(0,0)$. Therefore you get as the $\KER T=\{(1,0)\}$. $\DeclareMathOperator{\IM}{Im}\DeclareMathOperator{\KER}{Ker}$

Answer 2

You showed that $A\begin{pmatrix}x\\ y\end{pmatrix}=-\begin{pmatrix}x\\ y\end{pmatrix}$ if and only if $y=0$.

Thus the set of eigenvectors for the eigenvalue $-1$ is $\{(x,0)\mid x\ne0\}$ and the eigenspace for the eigenvalue $-1$ is $E_{-1}=\{(x,0)\mid x\in\mathbb R\}$. Note that $E_{-1}$ is the subspace generated by the vector $(1,0)$.