Eigenvalues of the transpose

I have tried this problem and I'm unsuccessful.

Let $T: M(n,n)\rightarrow M(n,n)$ be the mapping defined by $ T(A) = A^T$. Show that the only eigenvalues of $T$ are $\pm 1$.

@JohannesHahn has the best answer but here's something a bit more explicit.

Your map is $T:M_{n}\to M_n$ defined by $T(A)=A^\top$. An eigenvector of $T$ is a nonzero $n\times n$ matrix $A$ such that $A^\top=\lambda\cdot A$. Since $A$ is nonzero, it has a nonzero entry $a_{ij}$. But the equation $A^\top=\lambda\cdot A$ implies $$ a_{ji}=\lambda a_{ij}\tag{1} $$ and $$ a_{ij}=\lambda\cdot a_{ji}\tag{2} $$ Now, plugging (1) into (2) gives $$ a_{ij}=\lambda^2 a_{ij} $$ so $\lambda^2=1$.

Note that taking the transpose is a linear map of order two, that is $(A^T)^T=A$ for all $A$. Now every endomorphism of any vector space that satisfies $\phi^2=id$ can only have $\pm 1$ as eigenvalues. If $n>1$ then you can easily explicitely write down matrices that are eigenvectors to +1 and -1 respectively.

Hint: For a non-zero diagonal matrix the eigenvalue is clearly 1. If you have a non-zero off-diagonal element $a_{ij}$ and $a_{ji}$ then for the eigenvalue $\lambda$ you have $a_{ji} = \lambda a_{ij}$ and similarly $a_{ij} = \lambda a_{ji}$. So $\lambda^2 = 1$.