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 ?
Best Answer
Your solution is right: $x=z$ means that every vector which is solution to your system has to be of the form $$ v=\begin{pmatrix} x\\ y \\ x\end{pmatrix} $$ where $x,y$ are now "free" coefficients in the field of the vector space, so you can write that as $$ v=x\begin{pmatrix}1\\ 0 \\ 1\end{pmatrix}+y\begin{pmatrix}0\\ 1\\ 0\end{pmatrix} $$ This means that the vector space of the solutions (that is the eigenspace $E_4$) is spanned by the set $$ \left\{\begin{pmatrix}1\\ 0 \\ 1\end{pmatrix},\ \begin{pmatrix}0\\ 1\\ 0\end{pmatrix}\right\}. $$