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Prove or disprove:

if $A^2 \in M_{3}(\mathbb{R})$ is diagonalizable then so is $A$.

I'm pretty confident this is not true, but I've tried and tried to find a counter example without success. If someone contradicts this, I'd appreciate if you can outline your thought process when constructing the matrix $A$.

Also, another question in the same batch, asks:

Let $p(t)=t(t-0.25)(t-1)$ be the characteristic polynomial of $A^2$, is $A$ diagonalizable?

I'm thinking this is suppose to answer the previous question, assuming the answer here is false.

Here I know $A^2$ is diagonalizable, but I haven't made any substantial progress other than that.
I know I can go from the eigenvalues of $A$ to the eigenvalues of $A^k$, but not the other way around.

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Daniel, I have changed the notation to $M_{3}(\mathbb{R})$ which is the standard notation for set of all $3 \times 3$ real matrices. I hope that isn't a problem. If it is, then please rollback. –  user9413 Jul 6 '11 at 15:41
@Chandru, not at all, thank you. –  daniel.jackson Jul 6 '11 at 16:00
This is not true for matrices with a nilpotent Jordan component. However, if $A$ is non-singular, and we switch the field to complex numbers, then it is true. A slightly more general case (some power known to be diagonalizable) of this was recently discussed in fora.xkcd.com/viewtopic.php?f=17&t=71107 –  Jyrki Lahtonen Jul 6 '11 at 16:10
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4 Answers 4

up vote 8 down vote accepted

Try $A=\begin{pmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0\end{pmatrix}$. The idea is that the null matrix is diagonalizable, but there exist matrices which satisfy $A^2=0$ and $A$ is not diagonalizable.

For the second question, note that the eigenvalues of $A^2$ are distinct, and therefore the eigenvalues of $A$ are distinct, which implies that $A$ is diagonalizable.

If $P_A$ denotes the minimal polynomial of $A$ then $P_{A^2}(X^2)$ is an annulating polynomial for $A$. This means that $P_A|P_{A^2}(X^2)$. If $A^2$ is diagonalizable then $P_{A^2}$ has simple roots. If zero is not root of $P_{A^2}$ then all roots of $P_{A^2}(X^2)$ are simple so all roots of $P_A$ are simple and $A$ is diagonizable. (here we need to be sure that the roots of $P_A$ are all in the field of elements of $A$; if we work on $\Bbb{C}$ then we are fine)

So if zero is not an eigenvalue of $A$ then $A^2$ diagonizable implies $A$ diagonizable.

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The answer to the second problem is yes. In general, the (in general complex) eigenvalues of $A^2$ are the squares of the eigenvalues of $A$. In this case $A^2$ has the eigenvalues $0$, $\frac{1}{4}$, and $1$. Therefore $0$ is an eigenvalue of $A$, either $\frac{1}{2}$ or $-\frac{1}{2}$ is an eigenvalue of $A$, and either $1$ or $-1$ is an eigenvalue of $A$. This means that $A$ has $3$ distinct real eigenvalues, which implies it is diagonalizable. The key is that the eigenvalues of $A^2$ are distinct and nonnegative, which implies that the eigenvalues of $A$ are distinct and real.

Edit: I accidentally overlapped somewhat with Beni Bogosel's earlier answer. (More was added after I started writing.)

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So if $A^2$ had a negative eigenvalue, say $-1$, the same argument would not work? Because then it's possible that $A$ characteristic polynomial has $i$ as one of its roots, which is not an eigenvalue of $A$. –  daniel.jackson Jul 6 '11 at 16:04
@daniel.jackson: Yes, except you can strengthen your statements. If $A^2$ has a negative eigenvalue, then $A$ necessarily has a nonreal eigenvalue, which implies it is not diagonalizable in $M_n(\mathbb R)$. –  Jonas Meyer Jul 6 '11 at 16:07
Why must it have a nonreal eigenvalue? –  daniel.jackson Jul 6 '11 at 16:15
It follows from the fact mentioned above (of which I have not provided a proof) that the eigenvalues of $A^2$ are precisely the squares of the eigenvalues of $A$, while squares of real numbers are nonnegative real numbers. –  Jonas Meyer Jul 6 '11 at 16:21
I see. Could you be so kind to provide a proof (a general idea will be fine too)? –  daniel.jackson Jul 6 '11 at 16:26
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Try $A=\begin{pmatrix} 0&-1&0 \\ 1&0&0 \\ 0&0&1\end{pmatrix}$.
I found it by considering the complex matrix $\begin{pmatrix} 1&0&0 \\ 0&i&0 \\ 0&0&-i\end{pmatrix}$ and finding its characteristic polynomial, then using rational canonical form over $\mathbb{R}$.

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The idea here is that over $\mathbb{C}$, being diagonalizable is equivalent to having distinct eigenvalues. So, over $\mathbb{C}$, your question is true: $A^2$ diagonalizable implies $A$ diagonalizable. So to find a counterexample, you need a diagonal matrix over $\mathbb{C}$ whose eigenvalues are not all real, but whose squares are all real. –  user641 Jul 6 '11 at 15:54
uh? having distinct eigenvalues is sufficient but not necessary for being diagonalizable –  leonbloy Jul 6 '11 at 15:56
Ah, yes, sorry, I meant distinct roots in its minimal polynomial. –  user641 Jul 6 '11 at 15:57
No, $A^2$ diagonalizable does not imply $A$ diagonalizable. It doesn't matter what field this is over. See Beni Bogosel's answer. –  Robert Israel Jul 6 '11 at 16:48
I have been working with invertible matrices the whole time!! –  user641 Jul 6 '11 at 17:00
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To the second question: yes. If all eigenvalues of $A^2$ are pairwise distinct, then so are the eigenvalues of $A$, because if $Av=\lambda v$ then $A^2 v = \lambda ^2 v$, so $\ker(A-\lambda I) \subset \ker(A^2 - \lambda^2 I)$.

A matrix with pairwise distinct eigenvalues is diagonalizable. That's a simple and well known theorem.

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