Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This problem appeared in my algebra textbook

Suppose $A$ is a real matrix such that $A^2 = A^t.$ What are the possible eigenvalues of $A?$

Here is what I tried in order to solve it.

Since the eigenvalues of $A$ and $A^t$ are the same and since $A^2 = A^t$ it follows that $$\det(A-\lambda I) = 0 \iff \det(A-\sqrt{\lambda}I)\det(A+\sqrt{\lambda}I) = 0$$

So if $\lambda$ is an eigenvalue of $A$ then so is $\sqrt{\lambda}$, with appropriate +/- sign.

As far as I can see this implies that all eigenvalues of $A$ are either 0 or 1 since if any other $\lambda$ is an eignevalue of $A$ then so is $\sqrt{\lambda},\sqrt{\sqrt{\lambda}},\ldots$ which cannot possibly hapen for a matrix of finite size?

Is my reasoning correct?

share|cite|improve this question
This is not correct. Note that $\lambda$ need not be real even if $A$ is. It would be cleaner to say that if $\lambda$ is an eigenvalue so is $\lambda^2$ (and the converse). – Qiaochu Yuan May 22 '12 at 22:39
In other words: If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. But it is false to claim that if $\rho$ is an eigenvalue of $A^2$, then $\sqrt{\rho}$ is an eigenvalue of $A$: not only can $\sqrt{\rho}$ fail to be a real number, even if it is a real number you could have chosen the wrong sign. – Arturo Magidin May 23 '12 at 0:08
up vote 2 down vote accepted

Let $A^2=A^t$, then by taking transpose on both sides $(A^t)^2=A$, so $(A^2)^2=A$.

Now $A^4=A$.

Let $Ax=\lambda x$. with x one of the eigenvectors of A. Then $A^4x=\lambda x$. So $A^3 \lambda x=\lambda x$.Expanding like this, $\lambda^4x = \lambda x$. So $(\lambda^4-\lambda)x=0$. x being one of the eigenvectors of A is not all zero, whereas the other factor is just a constant. So we get $\lambda^4 - \lambda = 0$. In other words, $\lambda$ must be either zero or one of the cube roots of unity (including 1 itself). 0 is obviously attained by the zero matrix and 1 by any identity matrix. For the rest, consider

0   0   1
1   0   0
0   1   0

This satisfies the conditions in the question and its characteristic polynomial is $1-\lambda^3=0$. So its eigenvalues are the cube roots of unity.

share|cite|improve this answer
Very nice. I'm glad to see that it was possible to follow the steps in my answer. But you haven't left anything for OP to do. – Gerry Myerson May 23 '12 at 12:58
Yeah I should probably have mentioned that you had already done most of the stuff in your answer. I am not sure how to go about finding a good balance between helping the OP find the answer and not leaving anything to do, that comes from experience I guess. – Wonder May 23 '12 at 13:02

You are correct that the eigenvalues of $A^t$ and of $A$ are equal. That means that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ must be an eigenvalue of $A^2$, hence of $A^t$, hence of $A$. So if $\lambda$ is an eigenvalue of $A$, then so is $\lambda^2$, and hence $\lambda^4$, and hence $\lambda^8$, and hence $\lambda^{16}$ etc.

That means that the list $$\lambda,\lambda^2,\lambda^4,\lambda^8,\lambda^{16},\ldots,\lambda^{2^n},\ldots$$ must in fact be a finite list. This does not occur if $|\lambda|\gt 1$, because then the absolute value of these numbers is strictly increasing. It cannot happen if $0\lt|\lambda|\lt 1$ either, because then the absolute values of the terms of the sequence is strictly decreasing, so the list cannot be finite.

What does that leave?

P.S. This will reduce the possibilities to something tractable. You still need to figure out whether those possibilities can in fact occur.

share|cite|improve this answer
The finiteness of that list is certainly a necessary condition. It's not clear to me whether it is a sufficient condition. – Gerry Myerson May 23 '12 at 0:21
@GerryMyerson: You mean, whether the values that this consideration allows can actually occur? For two of them, it's easy; but there is indeed one case that seems difficult without bringing to bear some other ideas. – Arturo Magidin May 23 '12 at 2:02
I'm not quite with you. Does your "one case" cover an infinity of values of $\lambda$ that are consistent with the list being finite? – Gerry Myerson May 23 '12 at 4:23
@Gerry: I was thinking in terms of real eigenvalues only. – Arturo Magidin May 23 '12 at 4:30
Ah. Then we still have some work to do (if nonreal eigenvalues are permitted). – Gerry Myerson May 23 '12 at 6:24

You are given a relation between $A$ and $A^t$.

Take the transpose on both sides of that relation to get a second relation between $A$ and $A^t$.

Eliminate $A^t$ from the two relations to get an equation involving only $A$.

From that equation you should be able to reduce the eigenvalue possibilities to four.

Two of those four possibilities can certainly occur; I'm not sure about the other two.

share|cite|improve this answer
Consider $A$ in the block form $A = [z \; A ; 0 \; B]$ where $z$ is a complex number. Doesn't it follow from $A^2 = A^t$ that $z^2 = z$ and thus only $1,0$ are possible eigenvalues? Am I missing something crucial? – Jernej May 23 '12 at 10:56
@Azoo, sorry, I don't understand your block form, nor how you can write $A$ in it. – Gerry Myerson May 23 '12 at 12:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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