Find the eigenvalue and engenvector of a transformation $A$ on $\mathbb{R}^{2\times 2}$? Define an transformation $A$ on $\mathbb{R}^{2\times 2}$ by 
$$A(X)=MXN, \forall X\in {\mathbb R}^{2\times 2},$$
where $$
M = \pmatrix{1 & 0\\ 1 & 1}, \quad N = \pmatrix{1 & -1\\ -1 & 1}.
$$ Find the eigenvalues and engenvetors of the transformation $A$?
I wonder if we can consider $A$ on $\mathbb{R}^4$?
 A: Note: we have 
$$
M = \pmatrix{1 & 0\\ 1 & 1}, \quad N = \pmatrix{1 & -1\\ -1 & 1}.
$$
In your comment, you correctly calculate the matrix of this transformation relative to the basis $\mathcal B = \{E_{11},E_{21},E_{12},E_{22}\}$ (where $E_{ij}$ is the matrix with a $1$ in the $i,j$ entry and $0$'s elsewhere).  In particular, you found
$$
[A]_{\mathcal B} = N^T \otimes M = 
\pmatrix{1 & 0 & -1 & 0\\ 1 & 1 & -1 & -1\\ -1 & 0 & 1 & 0\\ -1 & -1 & 1 & 1}
$$
(where $\otimes$ denotes a Kronecker product). From there, we can calculate eigenvalues in the usual way to find that the eigenvalues are $2,2,0,0$, and proceed to calculate eigenvectors as usual.  The eigenvectors corresponding to $0$ are $(1,0,1,0)^T,(0,1,0,1)^T$, and the only eigenvector corresponding to $2$ is $(0,-1,0,1)$. Notably, our matrix (and our transformation) is not diagonalizable.  Converting these eigenvectors from their coordinate vectors to the corresponding matrices produces the eigenvalues/eigenvectors
$$
\lambda = 0: \quad \pmatrix{1&1\\0&0}, \pmatrix{0&0\\1&1}. \qquad 
\lambda = 2: \quad \pmatrix{0&0\\-1&1}.
$$
Note: the basis $\mathcal B$ chosen here is "special" in that the coordinate vector of a matrix relative to $\mathcal B$ corresponds to that matrix's vectorization.  It is for this reason that $[A]_{\mathcal B}$ has this nice formula.

An alternative approach is as follows: note that if $Mx = \lambda x$ and $N^Ty = \mu y$, then the transformation has the eigenvalue $\lambda \mu$ and corresponding eigenvector $X = xy^T$. Indeed, we find that
$$
A(X) = Mxy^TN = (Mx)(N^Ty)^T = (\lambda x)(\mu y)^T = \lambda \mu xy^T = \lambda \mu X.
$$
Since $M$ has $1$ as its only eigenvalue and $N$ has eigenvalues $0,2$, we find that $A$ has eigenvalues $0,2$. The vector $y_0 = (1,1)^T$ satisfies $N^Ty_0 = 0$. It follows that $X = xy_0^T$ satisfies $A(X) = 0$ for any $x \in \Bbb R^2$.  That gives us the eigenvalue $0$ and the corresponding eigenvectors 
$$
\pmatrix{1\\0}\pmatrix{1&1}, \quad \pmatrix{0\\1}\pmatrix{1&1}.
$$
The other eigenvector of $N$ is $2$. We note that $y_2 = (1,-1)^T$ satisfies $N^Ty_2 = 2y_2$.  Similarly, $x_1 = (0,1)$ is an eigenvector of $M$.  It follows that
$$
x_1 y_2^T = \pmatrix{0\\1} \pmatrix{1&-1}
$$
is an eigenvector of $A$ corresponding to $\lambda = 1$.
An unfortunate aspect of this approach, however, is that we still need to argue that there is no remaining eigenvector.

Yet another alternative (and absolutely absurd) approach to find the eigenvalues: We see that the row space of $A(X) = MXN$ must lie within the row space of $N$, which is to say that each row of $A(X)$ is necessarily a multiple of $(1,-1)$. So, the image of $A$ has dimension $2$, which means that $A$ has rank $2$. 
Thus, $A$ has $0$ as an eigenvalue with multiplicity $2$ (the dimension of the kernel of $A$) and non-zero eigenvalues $\lambda_1,\lambda_2$.  A convenient way to find these remaining eigenvalues is to consider the trace of $A$ and $A^2$.
We note that $\Bbb R^{2 \times 2}$ is an inner product space relative to the Frobenius inner product. By using the definition of trace on an inner product space, we find that 
$$
\operatorname{tr}_{\Bbb R^{2 \times 2}}(A) = \sum_{i,j = 1}^2 \langle E_{ij},A(E_{i,j} \rangle = \sum_{i,j = 1}^2 \operatorname{tr}_{\Bbb R^2}(E_{ij}^TA(E_{ij})) = \sum_{i,j = 1}^2[A(E_{ij})]_{i,j}.
$$
We compute
$$
[A(E_{11})]_{1,1} = \pmatrix{1&-1\\1&-1}_{1,1} = 1, \quad
[A(E_{12})]_{1,2} = \pmatrix{-1&1\\-1&1}_{1,2} = 1,\\
[A(E_{21})]_{2,1} = \pmatrix{0&0\\1&-1} = 1, \quad
[A(E_{22})]_{2,2} = \pmatrix{0&0\\-1&1}_{1,1} = 1. \quad
$$
Add these together to conclude that $\operatorname{tr}(A) = 4$.  Similarly, noting that $A^2(X) = M^2XN^2$, compute $\operatorname{tr}(A^2) = 8$.  Using the fact that the trace is the sum of the eigenvalues, we find that the eigenvalues $\lambda_1,\lambda_2$ must satisfy
$$
\lambda_1 + \lambda_2 = 4, \quad \lambda_1^2 + \lambda_2^2 = 8.
$$
Solving this system yields $\lambda_1 = \lambda_2 = 2$.
