2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $X^n$.

$\endgroup$
  • $\begingroup$ Thanks Elliot!! THis really answered my question! $\endgroup$ – user110320 Sep 2 '18 at 21:05
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.