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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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1  
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
1  
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 2

up vote 2 down vote accepted

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 :

Eigenvector when all terms in that column are zero?

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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