How to show $T$ satisfies $T^2=I$ if $T$ is such mapping that $T(A)=A^t$?

Question

Here how do you find the matrix of tranformation. I guess it is $A^{t-1}$. But how is $A^{2(t-1)}=I?$

In addition, let's say $\lambda =1,-1$ in this case (as I think so). How do you find eigenspace $E_{-1}$ and $E_{+1}$

Answer 1

You have $$ T^2(A) = T(T(A)) = T(A^t) = (A^t)^t = A $$