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Suppose $A$ is a square $n \times n (n \ge2)$ matrix such that $A^2 = A^T$. Prove that if $\lambda$ is a real eigenvalue of $A$ then either $\lambda = 0$ or $\lambda = 1$

Firstly I decided to find the determinant of $A$: $(\det A)^2 = \det A$ hence either $\det A = 0$ or $\det A = 1$. In the first case a zero eigenvalue obviously exists, so assume $\det A = 1$. In this case $A$ is invertible therefore $$AAAA = A^TA^T = (AA)^T = A$$ i.e. $$AAA^T = A$$ hence $$AA^T = I$$ So, $A$ is orthogonal. As I know eigenvalues of an orthogonal matrix can be $\pm1$

Could you please give me any hints how to prove that $\lambda = -1$ isn't possible?

Thanks in advance!

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  • $\begingroup$ I think this is a hint: $A$ and $A^T$ have the same eigenvalues $\endgroup$ – Tim kinsella Feb 26 '18 at 18:11
  • $\begingroup$ @Timkinsella Yes, I understand it, but unfortunately don't realize how to use this fact... $\endgroup$ – D F Feb 26 '18 at 18:16
  • $\begingroup$ Oh wait this might be a more useful fact: $A^4 = (A^2)^2 = (A^T)^2 = (A^2)^T= (A^T)^T=A$ $\endgroup$ – Tim kinsella Feb 26 '18 at 18:23
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If $\lambda$ is an eigenvalue of $A$ then we have$$|\lambda^2I-A^2|=|\lambda I-A||\lambda I+A|=0\\|\lambda I-A^t|=|(\lambda I-A)^t|=|\lambda I-A|=0$$which means that $\lambda$ and $\lambda^2$ are eigenvalues of $A^t$ and $A^2$ respectively. Since $$A^2=A^t$$ we have $$\lambda^2=\lambda$$ or $$\lambda=0\text{ or }1$$

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    $\begingroup$ I don't follow your argument. Why does $A^2=A^t$ imply that $\lambda^2=\lambda$? Where did you use the condition that $\lambda$ is real? Note that if $A=\pmatrix{c&-s\\ s&c}$ where $c$ and $s$ are respectively the cosine and sine of $2\pi/3$, we do have $A^2=A^t$, but none of the eigenvalues of $A$ satisfies that $\lambda^2=\lambda$. $\endgroup$ – user1551 Feb 26 '18 at 19:01
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$$AA^T=A(A^2)=A^3.$$

$$AA^T=(AA^T)^T=(A^T)^3=A^6.$$

So $0=A^6-A^3=A^3(A^3-1)$. The characteristic equation of $A$ must divide this expression, and hence the eigenvalues of $A$ must be within the set $\{0,1,e^{i\pi/3},e^{2i\pi/3}\}.$

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  • $\begingroup$ it's beautiful! Thank you $\endgroup$ – D F Feb 26 '18 at 18:34

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