# Let $A$ be a $n× n$ real matrix with $A^2 = A^T$. Show that every real eigenvalue of $A$ is either $0$ or $1$.

Let $A$ be a $n×n$ real matrix with $A^2 = A^T$. Show that every real eigenvalue of $A$ is either $0$ or $1$.

My thoughts:

$A^2 = A^T$
$\implies$ $A.A=A^T$
$\implies$$(A.A)^T=A \implies A^TA^T=A \implies$$A^2A^2=A$
$\implies$ $A^4-A=0$ .
so the real root of the equation $x^4-x=0$ are $0$ & $1$.
Am I right?

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you mean $A^2 = A^T$? – Learner Mar 15 '13 at 3:24
yes,of course.. – poton Mar 15 '13 at 3:25
That's great. You should write it as an answer. – 1015 Mar 15 '13 at 3:27

$A^2 = A^T$
$\implies$ $A.A=A^T$
$\implies$$(A.A)^T=A \implies A^TA^T=A \implies$$A^2A^2=A$
$\implies$ $A^4-A=0$ .
so the real root of the equation $x^4-x=0$ are $0$ & $1$.

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Conclude! The real roots of the polynomial $x^4-x$ which annihilates $A$ are $0$ and $1$. So the real eigenvalues of $A$ belong to $\{0,1\}$. Good job, +1. – 1015 Mar 15 '13 at 3:42

As pointed out by julien in his comment, you are going in the right direction. As $x^4-x$ is an annihilating polynomial of $A$, if $A$ has a real eigenvalue $\lambda$, then $\lambda$ must be a real root of $x^4-x$. Hence $\lambda=0$ or $1$.

Alternatively, you may prove the statement as follows. Let $(\lambda,v)$ be an eigenpair of $A$. By assumption, we have $v^TA^2v = v^TA^Tv$. However, since $A^2v=A(Av)=\lambda^2 v$ and $v^TA^T=(Av)^T=\lambda v^T$, the previous equation implies that $\lambda^2 v^Tv=\lambda v^Tv$. As $v$ is an eigenvector, $v^Tv=\|v\|^2\neq0$. Hence $\lambda^2=\lambda$, i.e. $\lambda=0$ or $1$.

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Nice. This is a much stronger result as the answer given by the OP as it rules out imaginary eigenvalues completely. – Elmar Zander Mar 15 '13 at 9:11
@ElmarZander No, it doesn't. I did have assumed that $\lambda$ is real. If this assumption wasn't made, what I got would be $\lambda^2=\lambda^\ast$ and we cannot rule out the existence of nonreal eigenvalues. In fact, nonreal eigenvalues can exist. Consider, e.g. the 2-by-2 matrix of rotation by $2\pi/3$. – user1551 Mar 15 '13 at 9:29
Oops. Yes, you're right. Stupid me ;-) – Elmar Zander Mar 15 '13 at 9:35

Let say eigen value of $A$ is $\lambda_i$.
We know that Eigen Value($A$) = Eigen value($A^T$).
Eigen value of $A^2$ are $\lambda_i^2$.
Now,Equating eigen value of $A^T$ and $A^2$

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Your argument is incomplete. All it shows is that the sets $\{\lambda_1,\ldots,\lambda_n\}$ and $\{\lambda_1^2,\ldots,\lambda_n^2\}$ are equal. You need to elaborate why you can immediately equate every individual $\lambda_i$ with $\lambda_i^2$ (e.g. by considering the complex Jordan form of $A$). – user1551 Mar 15 '13 at 4:19