3
$\begingroup$

Can you help me to find solution to this problem?

I have matrix $ A \in M_n(\mathbb{C}): A^2 + A^4 = 0$ and I know, that maximal number of linear independent eigenvectors is $4$. Also known, that minimal polynomial not equal to characteristic polynomial.

What we can say about Jordan canonical form of this matrix $ A$?

$\endgroup$
  • 1
    $\begingroup$ Based on the information you have, you should be able to explicitly write down the minimal polynomial. After this, list everything you know about the relationship between the minimal polynomial and the jordan canonical form (there is a certain theorem which allows you to construct the JCF once you know the minimal polynomial). $\endgroup$ – peek-a-boo Dec 23 '19 at 21:03
  • $\begingroup$ how i can explicitly write down the m.p.? It can be a polynomial with degree < 4 $\endgroup$ – Nickolay Kononov Dec 23 '19 at 21:12
  • $\begingroup$ There are many possibilities here. The simplest is a zero matrix. Do you need all of them? Then look at what Jordan blocks are possible (at most 4) from the annihilating polynomial $x^2(x+i)(x-i)$. $\endgroup$ – A.Γ. Dec 23 '19 at 21:23
1
$\begingroup$

[FINAL EDIT: In this answer, I made the main mistake of assuming that $n=4$ (among a myriad of other mistakes that the comments point out). This answer still might be a good reference to a specific case to really think about what happens when $n \geq 4$, so I'll keep it up. But please refer to my other answer which accounts for the general case where $n$ is arbitrary.]

So I assume the problem is asking about all the possible Jordan Canonical forms that satisfy this equation.

To do this, first we need to find all the possible minimal polynomials, which are:

$$x, \ x+i, \ x-i, x^2, \ x^2+1, \ x(x+i), \ x(x-i), \ x(x^2+1), \ x^2(x-i), \ x^2(x+i), \ x^2(x^2+1).$$ (EDIT: Missed $x^2, x^2+1, x(x+i), x(x-i), x^2(x+i)$ earlier)

NOTE: I did this by noting that the possible eigenvalues are the zeros of the polynomial $x^2(x^2+1)$ equation $A^4+A^2=0$ and noting that the highest length of the possible Jordan Blocks are bounded by the multiplicities of the roots.

Next, we look at the possible Jordan Blocks of each minimal polynomial. I'll do the case where the minimal polynomial is $x(x^2+1)$, and let you the other cases (note that the degree one polynomial cases are trivial, and the $x^2(x^2+1)$ only has one possbility).

Note that for $x(x^2+1)$, there must be at least one $0$-block, $i$-block, and $-i$-block. That's three Jordan blocks of degree one, so the remaining block can possibly be one of those three degree one blocks (either a $0$, $i$, or $-i$ block). We end up with the following possible diagonal matrices (up to re-ordering): $$\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & 0 & 0 & -i \end{pmatrix} \quad \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & 0 & 0 & -i \end{pmatrix} \quad \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & 0 & 0 & -i \end{pmatrix}$$

Again, I'm leaving the possibilities where the polynomial is different from $x(x^2+1)$ to you to figure out. Hope that helps.

EDIT 2: I'll also do the case where $x^2(x-i)$ is the minimal polynomial, so that you can see a case where not all the matrices are diagonal. Note such a matrix contains at least a $0$-block of length $2$ and an $i$-block of length one. One remaining dimension of the eigenspace is free and can make up either an $0$-black or $i$-block of length one. We find the possibilities here are: $$\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \end{pmatrix} \quad \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & 0 & 0 & i \end{pmatrix}.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Polynomials $x^2$, $x^2+1$, $x^2(x+i)$ etc can also be minimal here. Besides, Jordan form does not have to be diagonal (taking into account the double root for $x^2$). $\endgroup$ – A.Γ. Dec 23 '19 at 21:30
  • $\begingroup$ Point taken on missing polynomials (and edited accordingly). As for your second point, yes, Jordan form doesn't HAVE to be diagonal, but the case where the minimal polynomial is $x(x^2+1)$, all the possibilities HAPPEN to all be diagonal. I'll edit in another case where you end up with a 2-block possibility, since seeing such a case would be helpful to the questioner. $\endgroup$ – Dark Logician Dec 23 '19 at 21:49
  • $\begingroup$ And what about approach when dim(V) > 4? $\endgroup$ – Nickolay Kononov Dec 23 '19 at 21:54
  • $\begingroup$ It's not too different from the case where $\text{dim}(V)=4$ (didn't catch the $n$ in $M_n(\mathbb{C})$, my bad). Hold up. Let me write a new answer. $\endgroup$ – Dark Logician Dec 23 '19 at 21:59
0
$\begingroup$

Clearly the possible linear factors of the minimal polynomial $m_A$ are $x$, $x-i$ and $x+i$.

  1. If $A$ is diagonalizable then $m_A$ splits into only linear factors so it can be one of these: $$x, x+i, x-i, x(x+i), x(x-i), (x+i)(x-i), x(x+i)(x-i)$$

    Since $m_A$ must be different from the characteristic polynomial, we must exclude all matrices with the characteristic polynomial equal to the above. These have Jordan form $$\begin{bmatrix} 0\end{bmatrix}, \begin{bmatrix} i\end{bmatrix}, \begin{bmatrix} -i\end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & i\end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & -i\end{bmatrix}, \begin{bmatrix} i & 0 \\ 0 & -i\end{bmatrix}, \begin{bmatrix} 0 & 0 & 0\\ 0 & i & 0 \\ 0 & 0 & -i\end{bmatrix}$$

    All other Jordan forms with $\le 4$ blocks which are diagonal matrices with entries in $\{0,i,-i\}$ are permitted.

  2. On the other hand, if $A$ is not diagonalizable then the minimal polynomial is one of these: $$x^2, x^2(x+i), x^2(x-i), x^2(x+i)(x-i)$$ so there is at least one $0$-block of size $2$, all other $0$-blocks are of size $\le 2$, and all other blocks (those are $i$-blocks and $(-i)$-blocks) are of size $1$, if present. We have to exclude those matrices of the above form with characteristic polynomials equal to some of the above: $$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & i\end{bmatrix}, \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & -i\end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & 0 & 0 & -i\end{bmatrix}$$ All other Jordan forms with $\le 4$ blocks of the above form are permitted.
| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Here is my second answer that I believe is a sufficient guide to a full classification of all Jordan Canonical forms for arbitrary dimension $n$.

So first we need to find the possible minimal polynomials, which are: \begin{align*} x, \ x+i, \ x−i, \ x^2, \ x^2+1, \ x(x+i), \ x(x−i), & \\ x(x^2+1), \ x^2(x−i), \ x^2(x+i), x^2(x^2+1). & \quad (1) \end{align*}

Next, note that for $n<4$, the minimal polynomial must have a degree equal or less than $n$. So for example, $n=0$ is trivial, and for $n=2$, we can only have $x, x+i, x−i, x^2, x^2+1, x(x+i), x(x−i)$.

For the case $n \geq 4$, all the possible minimal polynomials listed in $(1)$ are included. To determine the possibilities based on a given minimal polynomial for a dimension $n$, we include at least one block of each root of the multiplicity of the minimal polynomial, finding the rest of the possibilities is then a combinatorial exercise of finding all the possible combinations of extra Jordan blocks such that the sum of lengths of each Jordan block adds up to $n$.

As in my previous answer, I'll do the cases for $x(x^2+1)$ and $x^2(x-i)$, when $n \geq 3$.

For $x(x^2+1)$, we find that there must be at least one $0$-block, $i$-block, and $−i$-block of length one. The rest of the possible blocks can be any $0$-block $i$-block, $-i$-block of length one. Each of the possibilities are diagonal matrices of the form: $$\begin{pmatrix} 0 & & & & \\ & \underbrace{\ddots}_{n_1} & & & \\ & & 0 & & \\ & & & i & \\ & & & & \underbrace{\ddots}_{n_2} \\ & & & & & i \\ & & & & & & -i \\ & & & & & & & \underbrace{\ddots}_{n_3} \\ & & & & & & & & -i \end{pmatrix},$$ with $n_1, n_2, n_3 \geq 1$ $n_1+n_2+n_3=n$.

For $x^2(x-i)$, we find that there must be at least one $0$-block of length $2$, and one $i$-block of length $1$. And the rest can be any arrangements of $0$-blocks up to length $2$ and $i$-blocks up to length $1$. What we get is $$\begin{pmatrix} 0 & & & & \\ & \underbrace{\ddots}_{n_1} & & & \\ & & 0 & & \\ & & & \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & \\ & & & & \underbrace{\ddots}_{n_2} \\ & & & & & \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ & & & & & & i \\ & & & & & & & \underbrace{\ddots}_{n_3} \\ & & & & & & & & i \end{pmatrix},$$ with $n_1 \geq 0$ and $n_2, n_3 \geq 1$ such that $n_1+2n_2+n_3=n$.

Hope I didn't make any clumsy misreads, and I hope you can see how to replicate the other cases (which are of similar nature to the ones I did).

| cite | improve this answer | |
$\endgroup$

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.