Exercise about Matrix diagonalization Well I have to diagonalize this matrix :
$$
\begin{pmatrix} 5 & 0 & -1 \\ 1 & 4 & -1 \\ -1 & 0 & 5 \end{pmatrix}
$$
I find the polynome witch is $P=-(\lambda-4)^2(\lambda-6)$
Now I want to know eignevectors so I solve $AX=4X$ and $AX=6X$ with $X=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$but I have a problem with the first system ! 
In the correction they say "after an elementary calculus we have $E_4=Vect\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\right)$"
And I don't know how and why because with my system I just find that $x=z$. 
Can you explain to me, please ? 

You calculated correctly.  The set of all eigenvectors of eigenvalue $4$ is a $2$-dimensional subspace.  The condition $x = z$ is equivalent to all matrices of the form
$$
\begin{pmatrix} x \\ y \\ x \end{pmatrix} 
= \begin{pmatrix} x \\ 0 \\ x \end{pmatrix} 
+ \begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}
= x \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} 
+ y \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},
$$
where $x,y \in \Bbb{R}$.  There is your basis.
