$M = \left(\begin{array}{ccc}0 & -3 & -2 \\1 & 3 & 1 \\1 & 2 & 3\end{array}\right)$
I want to find a basis $B$ such that matrix for $M$ w.r.t $B$ has the form:
$\left(\begin{array}{ccc}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{array}\right)$
The eigenvalues for $M = 2,2,2$.
$M-2I = \left(\begin{array}{ccc}-2 & -3 & -2 \\1 & 1 & 1 \\1 & 2 & 1\end{array}\right)$.
I tried to find eigenvectors. This is what I came up with.
$(M-2I)v_1 = 0$, then $v_1$ = $(-1,0,1)^t$
And so forth but I either get a wrong answer or it's not invertible.
How do I find the generalized eigenvectors so M is in Jordan Normal Form?