I am given the characteristic polynomial $x^2(x^2-1)$ and am asked to find all possible jordan canonical forms. What I have so far is:
Possible elementary divisors are: 1) $x,x,(x+1),(x-1)$, 2) $x,x,(x+1)(x-1)$, 3) $x^2,(x+1)(x-1)$, and 4) $x^2(x+1)(x-1)$. I therefore got the possible Jordan forms as:
1)=2) \begin{matrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}
and
3)=4)
\begin{matrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}
I'm really unsure if these are correct, so any insight would be greatly appreciated!
Best Answer
They are correct. What the characteristic polynomial $$x^2(x-1)(x+1)$$tells you is that there are at least $3$ cages, of which the cages for eigenvalues $1$ and $-1$ have size $1$. This only leaves the eigenvalue $0$, for which you have $2$ options:
Of course, all this is done up to permutation, meaning that you can change the order of the eigenvalues on the diagonal and still get a Jordan form.