# 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 $$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

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 $X^n$.

• Thanks Elliot!! THis really answered my question! Sep 2, 2018 at 21:05

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.