Define $T: M_{n\times n}\to M_{n\times n}$ by $T(A):= A^t$. Note that $T$ is a linear transformation.
Show that $1$ and $-1$ are the only eigenvalues of $T$.
Let $\lambda$ denote an eigenvalue of $T$. Then $T(A) = \lambda A = A^t$. We know that if $A$ is a symmetric matrix, then $A = A^t$ and if $A$ is a skew symmetric matrix then $-A = A^t$, ie. the only eigenvalues of $T$ are $1$ and $-1$ for the cases where $A$ is symmetric and skew symmetric.
Do I need to show these are the only possible times that $\lambda A$ can equal $A^t$? If so, how can I show that there are no other cases?
Best Answer
if $\lambda$ is an eigen value and the corresponding eigenvector is $A \neq 0,$ then $$A^\top = \lambda A $$ taking transpose gives you $$A = \lambda A^\top=\lambda^2 A $$ which show you $$\lambda^2 = 1\implies \lambda = \pm 1. $$