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 $\pm 1 ?$

T is a linear operator. As I understood this: find $\lambda$ that: $$T(A) - \lambda A = 0$$ If A is symmetric $A = A^t$ then $\lambda = 1$, if anti-symmetric $A = -A^t$ then $\lambda = -1$.
In general case, there is no $\lambda$ that satisfy $a_{ij} = \lambda a_{ji}$ for all $a_{ij}$
