Eigenvectors of the Zero Matrix Given the following matrix: $ \begin{pmatrix}
   -1 & 0 \\
    0 & -1 \\ 
   \end{pmatrix} $.
I have to calculate the eigenvalues and eigenvectors for this matrix, and I have calculated that this matrix has an eigenvalue of $-1$ with multiplicity $2$
However, here is where my problem comes in:
To calculate the eigenvector, I need to use:
$$ \begin{pmatrix}
   -1-\lambda & 0 \\          
    0 & -1-\lambda\ \\ 
   \end{pmatrix} $$
Multiply it by
$$ \begin{pmatrix}
                               x \\
                               y \\
                               \end{pmatrix} $$
and set it equal to $$ \lambda\ \begin{pmatrix}
                               x \\
                               y \\
                               \end{pmatrix} $$
Using my value of $\lambda = -1$, I end up having the following: $ \begin{pmatrix}
   0 & 0 \\          
    0 & 0 \\ 
   \end{pmatrix} $
which equals $ \begin{pmatrix}
                               -x \\
                               -y \\
                               \end{pmatrix}. $
However, apparently I am meant to get an eigenvector of $  \begin{pmatrix}
                               1 \\
                               0 \\
                               \end{pmatrix} $. I have no idea where I am going wrong
 A: Actually you can read the eigenvalues and eigenvectors just by inspection. Notice that
$$
\begin{pmatrix}
-1&0\\
0&-1
\end{pmatrix}
=
-I
$$
Now think for a minute. This matrix effectively just multiplies the input by $-1$. What eigenvectors could it have? Recall that eigenvectors are special directions along which matrix multiplication acts just like multiplying the input by some scalar $\lambda$. Guess what! We already know that this matrix simply multiplies the input by $-1$, so any direction will do (every non-zero vector is an eigenvector of this matrix).  What about eigenvalues $\lambda$? Well, you've probably guessed it: $\lambda=-1$. 
Pick a pair of linearly independent vectors to describe the whole eigenspace and you are done with no calculation whatsoever. I would pick the simplest pair imaginable, that is:
$$
x_1=\begin{pmatrix}1\\0\end{pmatrix}
\qquad
x_2=\begin{pmatrix}0\\1\end{pmatrix}
$$
But really, you could have picked a different one, such as:
$$
x_1=\begin{pmatrix}1\\2\end{pmatrix}
\qquad
x_2=\begin{pmatrix}0\\5\end{pmatrix}
$$
It makes no difference as long as they are linearly independent.
A: User uraf's answer is supernal, but if you wanted to solve this more methodically, then you could solve $(A - (1)I)$x $=0$ traditionally, as explained in these analogous questions :
Eigenvector when all terms in that column are zero?
What to do with an empty column in the basis of the null space?
$0$x $= 0 \iff 0x + 0y = 0$, so any $x, y$ satisfy this equation. In particular, the elementary basis vectors do.
