Diagonalizing a 3×3 matrix

Eigenvalue $4$. Eigenvector $v_1 = (1,1,1)$.

Eigenspace for $\lambda=1$.
$$A - I = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}.$$
The kernel has dimension $2$, spanned by $v_2 = (1,-1,0)$ and $v_3 = (1,0,-1)$.

Diagonalization.
$$P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{pmatrix}, \qquad D = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
Since the columns of $P$ are linearly independent, $P$ is invertible and $A = PDP^{-1}$.

Conclusion. $A$ is diagonalizable. Moreover,
$$A^k = P\begin{pmatrix} 4^k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}P^{-1}.$$