<s>[INST] Diagonalizability of a matrix 
Show that $$ A :=\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 2 \\ -2 & 0 & 3  \end{pmatrix}$$ is diagonalizable.

What I did:
First, I determined the characteristic polynomial $$\chi_A(X) = \det(X \cdot E_3-A)=(X-3)(X-1)(X-1)=X^3-5X^2+7X-3,$$
so the eigenvalues are $3$ and $1$.
I then determined the eigenspaces of each eigenvalue:
$$X=3: \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0  & 0 \\
-2 & 2 & 2  & 0 \\
-2 & 0 & 0  & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0  & 0 \\
0 & 2 & 2  & 0 \\
0 & 0 & 0  & 0 \\
\end{array}\right),$$ so $x_1=0$, $x_2=-x_3$, $x_3=x_3$ and thus $V_3(C) = \left< \begin{pmatrix} 0\\-1\\1 \end{pmatrix} \right>$.
Analogous:
$$X=1: \left(\begin{array}{@{}ccc|c@{}}
0 & 0 & 0  & 0 \\
-2 & 0 & 2  & 0 \\
-2 & 0 & -2  & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
-2 & 0 & 2  & 0 \\
0 & 0 & -4  & 0 \\
0 & 0 & 0  & 0 \\
\end{array}\right),$$ so $x_1=x_3=0$, $x_2=x_2$ and thus $V_1(A) = \left< \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$.
It now follows that $\dim(V_3(C)) + \dim(V_1(A)) = 1+1=2 \lt 3 = \dim(A)$ and because of the $\lt$, A shouldn't be diagonalizable, but it is.
So where's the mistake? Thanks in advance!
 [/INST] Your error lies in the computation of the eigenspace corresponding to the eigenvalue $1$, which is $2$-dimensional; it is spanned by $(1,0,1)$ and by $(0,1,0)$.
</s>