I have this matrix:
$$ \begin{pmatrix} 2 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 3 & -1 \\ \end{pmatrix} $$
I found the eigenvalues which are $2$ and $-3$.
So I substituted $2$ in the matrix:
$$ \begin{matrix} 0 & 1 & 0 \\ 0 & -2 & 2 \\ 0 & 3 & -3 \\ \end{matrix} $$
The sytem is:
$$ \left\{ \begin{array}{c} v_2=0 \\ -2v_2 +2v_3=0 \\ 3v_2-3v_3=0 \end{array} \right. $$
Which becomes...
$$ \left\{ \begin{array}{c} v_2=0 \\ v_2=v_3 \\ v_2=v_3 \end{array} \right. $$
So I suppose that the bases of the eigenspace of the eigenvalue $2$ is $(0,0,0)$, is that possible?
Eigenspace for the eigenvalue $-3$:
$$ \begin{pmatrix} 5 & 1 & 0 \\ 0 & 3 & 2 \\ 0 & 3 & 2 \\ \end{pmatrix} $$
$$ \left\{ \begin{array}{c} 5v_1+v_2=0 \\ 3v_2+2v_3=0 \\ 3v_2+2v_3=0 \end{array} \right. $$
$$ \left\{ \begin{array}{c} v_1=-(v_2)/5 \\ v_2=-(2/3) v_3 \\ v_3=-(3/2)v_2 \end{array} \right. $$
So the base I found is $(-1/5, -2/3, -3/2)$
No, it is not possible; remember that by the definition of eigenvectors we are looking for no trivial solution of $(A-2I)x=0$.
Indeed note that from the system you obtain $v_1\neq0,\;$ thus an eigenvector is $\;(1,0,0),\;$ as you can directly verify from the $(A-2I)$ matrix.