Diagonalizing a 3×3 matrix

Eigenspaces.
- $E_4 = \ker(A-4I)$ has dimension $1$, spanned by $v_1=(1,1,1)$.
- $E_1 = \ker(A-I)$ has dimension $2$, spanned by $v_2=(1,-1,0)$ and $v_3=(1,0,-1)$.

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

Conclusion. $A$ is diagonalizable with eigenvalues $4,1,1$.