Consider the matrix
$$ A = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 2 & 2 \\ -1 & 0 & 1 \end{pmatrix} \in \text{Mat}_3(\mathbb{R}) $$
Then I have to determine an invertible matrix $S \in \operatorname{Mat}_3(\mathbb{R})$ such that $S^{-1}AS$ is a diagonal matrix.
My attempt
If I have understood what I have to do correctly I need to find the eigenvectors for the matrix $A$. In an earlier question, I calculated the eigenvalues which gave me that $$ t_1 = t_2 = 0 \ \text{and} \ t_3 =2 $$ and afterwards the eigenspaces in order to calculate the geometric multiplicity which gave me the following basis
$$\operatorname{Span}(1,-2,1)^T $$ and $$ \operatorname{Span}(-1,1,1)^T $$ but where will the last basis come from that I need to have in order to construct my matrix $S$? I thought that as $t_1 = t_2 = 0$ they would have the same basis but if I construct $$ S = \begin{pmatrix} 1 & 1 & -1 \\ -2 & -2 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$ and afterwards find the inverse of $S$ and calculate $S^{-1}AS$ I don't get the desired result as it should be $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} $$ if I am not mistaken. I am not sure where my mistake is. Can you help me in the right direction?
I have examined the matrix, it has eigenvalues $2,2, 0$, rather than $0,0,2$.
Let's compute $A-2I=\begin{bmatrix} -1 & 0 & -1 \\ 2 & 0 & 2\\ -1 & 0 & -1\end{bmatrix}$. Rank of $A-2I$ is $1$, hence, there are two independent eigenvectors corresponding to $2$, that is the geometric multiplicty is $2$. Obviously $\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$ is an eigenvector for eigenvalue 2. Use the eigenvectors to construct $S$.
I will let you take it from here.