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 $\pm 1 ?$
 A: 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 $\lambda=-1,1$.
A: 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 $a_{ij}$
A: What you are trying to show is that if there is an $n\times n$ real matrix $A$ (not equal to the zero matrix) such that $T(A)=\lambda A$ for some scalar $\lambda,$ then $\lambda=1$ or $\lambda=-1.$  Moreover, you are trying to show that $T$ has eigenvalues of $\pm 1,$ meaning that you need to exhibit some particular matrices $A_1,A_2$ of the kind I mentioned above, such that $T(A_1)=A_1$ and $T(A_2)=-A_2.$ (The matrix in your question gives you a start on finding such a matrix $A_1$. I leave it to you to find such an $A_2.$)
To prove these are the only ones, suppose $T(A)=\lambda A$ and consider $$T\bigl(T(A)\bigr).$$
