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
Best Answer
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.