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