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 $\pm 1$.
 A: @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 $\lambda^2=1$.
A: 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.
A: 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 $\lambda^2 = 1$.
A: The map $T$ that sends a matrix to its transpose is a linear map between vector spaces. This means that once you choose a basis you can write $T$ as a linear map and then find its eigenvalues.
