Finding Jordan canonical form of a matrix given the characteristic polynomial I am trying to find the Jordan canonical form of a matrix $A$ given its characteristic polynomial. Suppose $A$ is a complex $5\times 5$ matrix with minimal polynomial $X^5-X^3$. The end goal of the problem is to find the characteristic polynomial of $A^2$ and the minimal polynomial of $A^2$. 
I know that since the minimal polynomial of a matrix divides the characteristic polynomial of a matrix, then $A$ has the same minimal and characteristic polynomial, namely $X^5-X^3$. Now I am trying to find the JCF (Jordan Canonical Form) of $A$ to make it easier to compute $A^2$, since $A$ is conjugate to its JCF. So, since the characteristic polynomial of $A$ splits into $X^3(X+1)(X-1)$, then I know that the Jordan canonical form will have three Jordan blocks, 2 of size 1 corresponding to $1$ and $-1$ and one of size 3. Now, my problem is that I can't figure out the form of this third Jordan block. How do I know that it has the from $$\begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & 0
\end{bmatrix}$$
or the form 
$$ \begin{bmatrix}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    0 & 0 & 0 
\end{bmatrix}$$
or the form 
$$\begin{bmatrix}
    0 & 1 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & 0
\end{bmatrix}$$
Thanks for all your help!
 A: We have the minimal polynomial is $X^3(X^2-1)$. Over $\Bbb C$, the exponent of the irreducible factor $(x-a)$ in the minimal polynomial gives the size of the largest Jodan block. Thus we have a $3\times 3$ Jordan block corresponding to the eigenvalue $0$. The only possibility is the middle one:
$$A=\left(
\begin{array}{ccccc}
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right).$$
To see this, note that the other two cases you gave have smaller minimal polynomials. For example, if
$$A=\left(
\begin{array}{ccccc}
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right),$$
then $A$ is annihilated by $X^2(X^2-1)$. If its not clear why, notice that an $n\times n$ matrix with all zeros and ones on the super diagonal is nilpotent with minimal polynomial $X^n$.
A: The minimal polynomial is $m_A(x) = x^3(x-1)(x+1)$ and the characteristic polynomial is the same up to sign change $\chi_A(x) = -x^3(x-1)(x+1)$. 
We can assume that $A$ is equal to its Jordan form, which is:
$$A = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}$$
We know that the $0$-block is of size $3 \times 3$ because $x^3$ is its power in $m_A$.
Squaring gives:
$$A^2 = \begin{bmatrix}
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}$$
Hence the characteristic polynomial is $\chi_{A^2}(x) = -x^3(x-2)^2$. The minimal polynomial can be seen to be $m_{A^2}(x) = x^2(x-1)$ because squaring the above matrix annihilates the $0$-block, and deducting $1$ annihilates the $1$-block.
