I am not sure how to go about solving this question.
We are given
$$A =
\left[
\begin{array}{ccc}
-1& 0& -2 \\
2& -3& -2 \\
0 & 0& -3\\
\end{array}
\right]
$$
and it has the eigenvalues -3, of multiplicity 2, and -1, of multiplicity 1.
How would I go about finding an orthogonal basis for the eigenspace corresponding to the eigenvalue -3? The answer is a span of 2 vectors
Row reducing A-3I just gives the identity matrix, and from there I'm stuck.
Best Answer
Since the eigenvalue is $-3$, what matters here is$$A+3\operatorname{Id}=\begin{bmatrix}2&0&-2\\2&0&-2\\0&0&0\end{bmatrix}.$$It is clear that, for instance, a basis of $\ker(A+3\operatorname{Id})$ is $\bigl\{(1,0,1),(0,1,0)\bigr\}$. So, an orthogonal basis of the eigenspace corresponding to $-3$ would be$$\left\{\left(\frac1{\sqrt2},0,\frac1{\sqrt2}\right),(0,1,0)\right\}.$$