Eigenvalues of such matrix $A^2 = A^T$

Question

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!

Answer 1

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 $$\lambda=0\text{ or }1$$

Answer 2

$$AA^T=A(A^2)=A^3.$$

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

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 $\{0,1,e^{i\pi/3},e^{2i\pi/3}\}.$