For the matrix
$$A = \begin{bmatrix} −1 & 0 & 0 \\ 3 & −1 & −3 \\ −3 & 0 & 2 \end{bmatrix}$$
the eigenvalue $\lambda = −1$ has a two-dimensional eigenspace. Find a basis for this eigenspace.
Is this right?
$$ -I – A = \begin{vmatrix} 0 & 0 & 0 \\ -3 & 0 & 3 \\ 3 & 0 & -3 \end{vmatrix}$$
$$ -> \begin{vmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{vmatrix}$$
let $x_3 = t, x_2 = s, x_1 = t$
So:
$$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} t \\ s \\ t \end{bmatrix} = t * \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + s * \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$
So the basis is $$(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix})$$
Best Answer
Yes, the solution is correct. There is an easy way to check it by the way. Just check that the vectors $\begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix}$ and $\begin{pmatrix}0\\\ 1 \\\ 0 \end{pmatrix}$ really belong to the eigenspace of $-1$. It is also clear that they are linearly independent, so they form a basis. (as you know the dimension is $2$)