Find eigenvalues for $f(A) = A^t$

Question

Let $f(A)=A^t$ be a linear transformation. Prove that its eigenvalues are $-1$ and $1$.

My try:

$f^2(A) = A$ So, $\forall A,f^2(A)=1 \cdot A$ . That means that $1$ is a solution of $P_A(x)$, so its also a solution of $P_{A^t}(x) \implies $ $1$ is an eigenvalue of $f$. I don't know what to do abut $-1$.

Answer 1

As you said, any eigenvalue of $f$ is necessarily a square root of $1$, so either $1$ or $-1$. To check that these are both actually represented, you should give a matrix $A$ with $A^T=A$ and another matrix $B$ with $B^T=-B$. (Some assumptions you haven't given us are required to ensure $B$ exists, but $A$ always exists (consider for instance $A=I$)).

Answer 2

minimal polynomial of f divides x^2-1, and hence -1 and 1 are the only possible eigenvalues of f and it is indeed the case. Take A to be symmetric matrix for 1 and Skew symmetric for -1.

Answer 3

You are almost there as $f^2(A)=A$ is just $(f^2-I)(A)=0$ corresponding multiple of minimal polynomial will be $(x^2-1)$ which clearly implies that $(x+1)(x-1)=0$ and hence the result.

Answer 4

Every $A$ may be written as $$ A = \frac{1}{2}(A+A^t)+\frac{1}{2}(A-A^t), $$ and $$ f(A) = \frac{1}{2}(A+A^t)-\frac{1}{2}(A-A^t). $$ So $A$ has only $\pm 1$ as possible eigenvalues. Symmetric matrices are eigenvectors with eigenvalue $+1$; anti-symmetric matrices are eigenvectors with eigenvalue $-1$. There are no other eigenvalues because $f^2=I$.