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If $A$ be a matrix with real entries and $A^2=A^t$, where $A^t$ denotes transpose of $A$, then the only real eigenvalues of $A$ are $0$ and $1$

I solved this problem using property that trace of a matrix is the sum of its eigenvalues. Then, I compared the trace of $A^2$ with $A^t$ which is equal to $A$. Then, I used Cauchy-Schwarz inequality on the eigenvalues and used the fact that square of a real number is always nonnegative to obtain the result.Am i right in that? I think we can also prove it using Jordan decomposition. Any ideas. Thanks beforehand.

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You could also have done like this:

$ A={(A^t)}^{t}={(A^2)}^{t}={(A^t)}^{2}={(A^2)}^{2}=A^4 $

So the polynomial:

$P=X^4-X=X(X-1)(X^2+X+1)$ Cancels $A$

We hence conclude that:

$sp(A)$ ($sp(A)$ is the set of eigenvalues of $A$) is included in the roots of $P$, and the real roots of $P$ are $0$ and $1$. It follows that the only real roots that $A$ CAN have are $0$ and $1$

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  • $\begingroup$ This is a nice answer; but just one comment. The complex roots of $X^2+X+1=0$ can arise as eigenvalues of real symmetric matrices with $A^2=A^t$. $\endgroup$ – Lord Shark the Unknown May 13 '17 at 10:40
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    $\begingroup$ if $A$ is a real symmetric matrix, then, the only eigenvalues of $A$ would be $0$ and $1$, besides, the question only asks for eventual real eigenvalues, so, it's not a problem having complex eigenvalues ^^ $\endgroup$ – Oussama Boussif May 13 '17 at 10:51

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