Eigen values of a transpose operator Let $T$  be linear operator on $M_{nxn}(R)$ defined by $T(A)=A^t.$ 
Then $\pm 1$ is the only eigen value.
My try : 
Let $n=2,$ then $[T]_{\beta}$ = $ 
\begin{pmatrix}
  a & 0 & 0 & 0 \\
  0 & c & 0 & 0 \\
  0 & 0 & b & 0 \\
  0 & 0 & 0 & d 
\end{pmatrix}$
Then the corresponding eigen values are $\lambda = a,b,c,d  $
How come i can claim it's eigen value is $\pm 1 ?$

If $T(A)=\lambda A$, then $T^2(A)=\lambda^2 A$.  But $T^2=I$ which only has an eigenvalue of 1.  So $\lambda^2=1$ or $\lambda=-1,1$.
