# 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

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As you've noted, you need to set $AX=\lambda X$, but you've done $(A-\lambda I)X=\lambda X$. –  Git Gud May 19 at 22:55
Tip: Put the periods before the ending $$. – Arkamis May 19 at 23:04 @Amzoti - surely my value of λ ensures that any v_i satisfies [A−λI]vi=0? – user136650 May 19 at 23:07 ## 2 Answers Actually you can read the eigenvalues and eigenvectors just by inspection. Notice that$$ \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.

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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 :

What to do with an empty column in the basis of the null space?

$\mathbf{0x = 0} \iff 0x + 0y = 0$, so any $x, y$ satisfy this equation. In particular, the elementary basis vectors do.

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