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!