Checking if matrix is diagonalizable

By using $\dim V_C(λ)=\dim(\Bbb R^{3\times3})−r(C−λI)$ we get that $\dim V_C(0)=2$ and $\dim V_C(3)=1$ so it is diagonalizable. Now from the solutions of the two systems we find a basis of $V_C(0)\setminus \{0\}$, say $$\begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\-1\end{pmatrix}\;,$$ and a basis of $V_C(3)\setminus\{0\}$, say $$\begin{pmatrix}1\\1\\1\end{pmatrix}\;.$$ These three vectors are a basis of $\Bbb R^{3\times3}$