I don't believe this matrix is diagonalizable, but I wanted to make sure.
Here's the matrix:
$A=\begin{bmatrix} 4 & 0 & 0 \\2 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$
A nice thing about this matrix is that it is diagonal, so the eigenvectors are $\lambda_1 = 4$, and $\lambda_2 = 2$ with algebraic multiplicity $2$.
The associated eigenvector for $\lambda_1$ is $x_1 = (1,1,1)$. When trying to solve for the eigenvector of $\lambda_2$, one gets the reduced matrix
$C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Here, there is only the trivial solution, and that can't be an eigenvector due to definition. Based on that, can I rightly say that this matrix is not diagonalizable?