Finding the Eigenvectors given Eigenvalues

Question

My matrix is

$$A=\begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}$$

I used $$det(\lambda I-A)=0$$

And found my two eigenvalues, $1$ and $-1$.

If I from here try to use $\lambda = -1$ it results in the equations

$$-v_1+v_2=0$$ $$v_1-v_2=0$$

Am I then correct in that I can from this point just state that $$v_1=v_2$$ and therefore the eigenvector can be expressed as $$\begin{bmatrix}v_1 \\v_1\end{bmatrix}$$ where I can just pick a value for $v_1$?

Also, how would you express this answer concisely (if my work is correct that is)?

Answer 1

It's correct. You could express the entire eigenspace as $E_{-1}=\{t\cdot\begin{pmatrix}1\\1\end{pmatrix}\mid t\in \Bbb R\}$.

Answer 2

You know that $(A-I)(A+I)=(A+I)(A-I)=0$. So the column vectors of $A+I$ are eigenvectors with eigenvalue $-1$. And the column vectors of $A-I$ are eigenvectors with eigenvalue $1$.

The column space of $A-I$ is one-dimensional: $$ A+I = \left[\begin{array}{cc}-1 & -1 \\ -1 & -1\end{array}\right]. $$ The column space of $A+I$ is one-dimensional: $$ A-I = \left[\begin{array}{cc}1 & -1 \\ -1 & 1 \end{array}\right]. $$

This is another way to get at what you are doing.