Find basis for the eigenspace of the eigenvalue

Question

We have the matrix $$\begin{pmatrix} 3 & 2 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -3 \end{pmatrix}$$

We want to find a basis for the eigenspace of each eigenvalue of this matrix. This matrix has three eigenvalues, $\lambda = 1, 2, -3$. For $\lambda = -3$, we have:

$$(A + 3I)\mathbf{v} = \begin{pmatrix} 6 & 2 & 0 \\ -1 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies \begin{cases} 6v_1 + 2v_2 = 0\\ -v_1 + 3v_2 = 0 \\ v_3 = ? \end{cases} \implies v_2 = -3v_1 \land v_3 = ?$$

So $(v_1, v_2, v_3) = (v_1, -3v_1, ?) = v_1(1,-3, ?)$ so $(1, -3, ?)$ is a basis for the eigenspace.

Now as you can see I have put $v_3 = ?$ everywhere because we don't know its value and I don't really know what to do there. However, a linear algebra tool tells me that the basis for the eigenspace (or well, at least the eigenvector) is $(0,0,1)$, which is totally different from my answer. So what am I doing wrong?

Answer 1

From $-v_1+3v_2=0,$ we conclude that $v_1=3v_2.$ Since $6v_1+2v_2=0,$ we then have $20v_2=0$ by substitution, so $v_2=0,$ and so $v_1=0.$

Now, your third equation shouldn't be $v_3=?,$ but rather $0=0.$ This is always true! Hence, all we need is to make $v_1=v_2=0,$ and we are free to make $v_3$ be whatever we want! So, the eigenvectors associated to $-3$ are those of the form $(0,0,v_3).$ All we need for a basis, then, is a non-zero vector of this form.