Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$ I need help working out the eigenvectors for this matrix. 
$ \begin {pmatrix} 0 &  0 \\   0 & -7 \end{pmatrix} $ 
The original matrix is $ \begin {pmatrix} 5 &  0 \\   0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,
but I am not sure how to about the eigenvectors, 
as for 5 
$ \begin {pmatrix} 0 &  0 \\   0 & -7 \end{pmatrix} $ $ \begin{pmatrix} x \\ y \end{pmatrix}$ = $ \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
from the first equation, $x$ and $y$ are both zero, but from the second equation $y = 0$, so what is the eigenvector? 
 A: To find the eigenvectors, you need to solve the linear system :
$$(A-\lambda I)v = 0$$
For the case of $\lambda = 5$, you have : 
$$(A-5I)v_5=0 \Rightarrow\begin{bmatrix} 0 & 0 \\ 0 & -7 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow \{ 0v_1 + 0v_2 = 0 |v_2 = 0 \}$$
This means that the eigenspace gets generated by $\{(v_1,v_2) \in \mathbb R^2 | v_1\in \mathbb R  \space \text{and} \space v_2=0\}$, thus $v_1$ can take any value over the reals since $0v_1 = 0$ is true for all $v_1 \in \mathbb R$. Simply letting $v_1=1$, you yield the eigenvector : 
$$v_5 = \begin{bmatrix}  1 \\ 0\end{bmatrix}$$
A: No, from the first equation, $x$ and $y$ are free. From the second equation, $y=0$. So your eigenvector is 
$$
\begin{bmatrix}1\\0 \end{bmatrix}
$$
as you can check, the equation is satisfied.
A: From first equation you deduce whatever is x and y the equation holds
$$0x+0y=0$$
From second equation you deduce that $y=0$
$$0x-7y=0 \implies -7y=0 \implies y=0$$
So 
$$(x,y)=(x,0)=x(1,0)$$
