Eigenvalues and eigenvectors of $A$ Find the eigenvalues and eigenvectors of linear transformation $A: M_{nxn} \to M_{nxn}$ defined as $A(X)=X^T$. 
I understand that $X$ looks like this:
\begin{bmatrix}
x_{11} && x_{12} && ... && x_{1n} \\ 
x_{21} && x_{22} && ... && x_{2n} \\ 
...    && ...    && ... && \\
x_{n1} && x_{n2} && ... && x_{nn} 
\end{bmatrix}
but I don't understand how I am supposed to proceed or get the solution. I would really appreciate some help, thanks.
I know a similar question has already been posted, but the answer there is not helpful to me. There the OP wanted to find the determinant, which could eventually lead me to my eigenvalues and eigenvectors, but I just don't understand their step by step procedure. 
 A: Really big hint: $A^2=I$  
So $\lambda^2=\dots$
A: We have that $X$ is an eigenvector of $A$ if, and only if, $X^T=\lambda X$. This is equivalent to saying that$$x_{ji}=\lambda x_{ij},\;\forall 1\le i,j\le n.$$ Now we observe that $$X=(X^T)^T=\lambda X^T.$$ This is equivalent to saying $$x_{ij}=\lambda x_{ji},\;\forall 1\le i,j\le n.$$ So now we have the following equality $$x_{ij}=\lambda ^2x_{ij},\;\forall 1\le i,j\le n.$$
Clearly this is only satisfied if either $\lambda=\pm1$ or $x_{ij}=0\;\forall 1\le i,y\le n$ however the latter results in $X=0$ which is not an eigenvector. Therefore we have that the only eigenvalues are $\lambda =\pm1$. $\lambda =1$ gives us $x_{ij}=x_{ji}\,\forall 1\le i,j\le n,$ i.e., $X$ (an eigenvector) is a symmetric matrix. $\lambda =-1$ forces $x_{ii}=0,\,\forall 1\le i\le n$ and $x_{ij}=-x_{ji},\,\forall 1\le i,j\le n,\,i\ne j$.

The explanation of one way to find the geometric multiplicities is as follows.  
First recall that the geometric multiplicity of an eigenvalue, $\lambda$, of $A$ is defined as the dimension of the $\lambda$-eigenspace of $A$, denoted $E_A(\lambda)$. If $X\in E_A(-1)$ then we know that the diagonal entries of $X$ must be zero. Observe also that if we choose an entry $x_{ij}$ of $X$ then this forces $x_{ji}=-x_{ij},\,i\ne j,$ as explained above. This leaves us with only $\frac{n(n-1)}{2}$ choices for the entries of $X$ and hence tells us (in a naïve sort of sense) that $$dim(E_A(-1))=\frac{n(n-1)}{2}.$$ (Check this for yourself).  
Now for $\lambda =1$. This is a very similar situation to $\lambda =-1$. If $X\in E_A(1)$ and if we make a choice for $x_{ij}$ then this forces $x_{ji}=x_{ij},\,i\ne j,\,$ (since $X$ is a symmetric matrix) and as shown this gives us $\frac{n(n-1)}{2}$ choices for non-diagonal entries of our matrix $X$. The difference here is that we are permitted non-zero entries on the diagonal of $X$ so this gives us a further $n$ choices for entries of $X$ and therefore we have that $$dim(E_A(1))=\frac{n(n-1)}{2}+n=\frac{n(n+1)}{2}.$$
A: If $i,j\in\{1,\ldots,n\}$, let $E_{ij}$ be the matrix such that the entry at the $i$-th column and $j$-th row is equal to $1$ and all other entries are equal to $0$. Consider the basis of $M_{n\times n}$ whose elements are the matrices $E_{i,i}$ as well as the matrices $E_{i,j}+E_{j,i}$ and $E_{ij}-E_{j,i}$ ($j\neq i$). Each vector of this basis is an eigenvector of $A$. To be more precise, $A\bigl(E_{i,i}\bigr)=A_{i,i}$, $A\bigl(E_{i,j}+E_{j,i}\bigr)=E_{j,i}+E_{i,j}=E_{i,j}+E_{j,i}$, and $A\bigl(E_{i,j}-E_{j,i}\bigr)=E_{j,i}-E_{i,j}=-\bigl(E_{i,j}-E_{j,i}\bigr)$.
