Can a diagonalizable matrix have entries that are not on the diagonal?

Question

Calculate the eigenvalues and eigenvectors of the real matrix

\begin{bmatrix} 2& 0 & 0 \\ 0 & 2& 0 \\ 1 & 1 &3 \end{bmatrix}

Is A diagonalizable? If so, find the matrix P by which you need to conjugate to produce a diagonal matrix

I have been trying to solve this problem first I worked out the eigenvalues as 2 and 3 and the eigenvectors as $(−1, 0, 1)^T, (0, −1, 1)^T$ for 2 and $(0,0,1)^T$ for 3.

Then I took these as a basis B and tried to work out $_{B}[A]_{B}={_{B}[id]_{e}}{_{e}[A]_{e}}{_{e}[id]_{B}}$ where e is the standard basis of $R^3$ but as this is meant to give $D=P^{-1}AP$, $_{B}[A]_{B}$ should give a diagonal matrix for D but I keep getting \begin{bmatrix} 3& 1 & 1 \\ 0 & 2& 0 \\ 1 & 0 &2 \end{bmatrix} Which is not a diagonaal matrix, where am I going wrong?

I also got P as \begin{bmatrix} 0& -1 & 0 \\ 0 & 0& -1 \\ 1 & 1 &1 \end{bmatrix} which is apparently correct.

Answer 1

$P=\begin{bmatrix} 0& -1 & 0 \\ 0 & 0& -1 \\ 1 & 1 &1 \end{bmatrix}$, and $A=\begin{bmatrix} 2& 0 & 0 \\ 0 & 2& 0 \\ 1 & 1 &3 \end{bmatrix}$.

So $P^{-1}=\begin{bmatrix} 1& 1 & 1 \\ -1 & 0& 0 \\ 0 & -1 &0 \end{bmatrix}$. Then you get:

$$P^{-1}AP=\begin{bmatrix} 3& 0 & 0 \\ 0 & 2& 0 \\ 0 & 0 &2 \end{bmatrix}$$