Find the eigenvalues and show the eigenvectors span

Question

Define $T : M_n(\mathbb{R}) \longrightarrow M_n(\mathbb{R})$ by $T(A) = A^t$. Here $M_n(\mathbb{R})$ denotes the set of $n\times n$ real matrices, and $A^t$ denotes the transpose of $A$. How would I prove that $T$ has only two distinct eigenvalues, and that its eigenvectors span $M_n(\mathbb{R})$?

Answer 1

If $A$ is an eigenvector of $T$ with eigenvalue $\lambda$, then\begin{align}A&=T\bigl(T(A)\bigr)\\&=T(\lambda A)\\&=\lambda T(A)\\&=\lambda(\lambda A)\\&=\lambda^2A\end{align}and therefore $\lambda^2=1$. So, $\lambda=\pm1$.

On the other hand, each matrix $A$ can be written as$$A=\frac12(A+A^t)+\frac12(A-A^t).\tag1$$The matrix $M=\frac12(A+A^t)$ is such that $M^t=M$ and the matrix $N=\frac12(A-A^t)$ is such that $N^t=-N$. So, $(1)$ expresses an arbitrary element of the space as the sum of two eigenvectors and therefore the eigenvectors span the whole space.