I am trying to diagonalise the following matrix $$ A = \begin{pmatrix} \sqrt{8} & 0 & 0 \\ 4 & \sqrt{8} & -4 \\ 0 & 0 & \sqrt{8} \end{pmatrix} $$ finding its eigenvalues.
The characteristic polynomial is
$$ (\lambda - \sqrt{8})^3 = 0 $$
therefore I have $\lambda = \sqrt{8}$ with algebraic multiplicity $3$. Now, getting to the eigenvectors I set $(A - \lambda I)\vec{v} = 0$ and this gives the equation
$$ 4v_1 - 4v_3 = 0 $$
therefore I conclude that I can only find no more than two independent vectors, for example $\vec{v}_1 = \begin{pmatrix} 1\\0\\1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} 0\\1\\0 \end{pmatrix}$ therefore the geometric multiplicity of the eigenvalue is $2$, then the matrix is not diagonalisable and we have to find the Jordan form.
Here is my problem.
When I try to find the generalised eigenvectors
$$ (A-\lambda I)^n\vec{v} = 0 $$
I find the null matrix. In fact, already
$$ (A-\lambda I)^2 = \begin{pmatrix} 0 & 0 & 0 \\ 4 & 0 & -4 \\ 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 0 & 0 \\ 4 & 0 & -4 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
therefore also for every $n$. How should I go on from here?
The eigenspace $E_{\sqrt8}$ corresponding to the eigenvalue $\sqrt8$ is the space spanned by $(1,0,1)$ and by $(0,1,0)$. In other words, it's the space of those vectors of $\Bbb R^3$ of the form $(a,b,a)$ ($a,b\in\Bbb R$).
On the other hand, if $(x,y,z)\in\Bbb R^3$, then$$A.(x,y,z)-\sqrt8(x,y,z)=\left(0,4x-4z,0\right)\tag1$$and this vector belongs to $E_{\sqrt8}$. So, take $(x,y,z)\in\Bbb R^3\setminus E_{\sqrt8}$. For instance, take $(x,y,z)=\left(\frac18,0,-\frac18\right)$, and then the RHS of $(1)$ is $(0,1,0)$. So, if
then $Av_1=\sqrt8v_1$, $Av_2=\sqrt8v_2$, and $Av_3=\sqrt8v_3+v_2$. So, if$$P=\begin{bmatrix}1&0&\frac18\\0&1&0\\1&0&-\frac18\end{bmatrix}$$(the columns of $P$ are the vectors $v_1$, $v_2$, and $v_3$), then$$P^{-1}AP=\begin{bmatrix}\sqrt8&0&0\\0&\sqrt8&1\\0&0&\sqrt8\end{bmatrix}.$$