Eigenvalues of a linear transformation $M_{2,2}\ \rightarrow M_{2,2}$ I took linear algebra years ago, and for various reasons I'm going back to re-learn it. In that process, I came across the following question in an old exam from the course I once took:
A linear operator g: $M_{2,2}\ \rightarrow M_{2,2}$ is given by:
$$ A\rightarrow  \pmatrix{0 & 1\\1 & 0} A - A^T \pmatrix{0 & 1\\1 & 0} $$
Determine the eigenvalues of g and the dimensions of the relative eigenspaces.
Looking over the textbook I used in that course, I didn't see any material on eigenvalues of general linear transformations, but I decided to give it a shot.
To do this, I set $$A=\pmatrix{a & b\\c & d} $$ and computed the right hand side, so:
$$ \pmatrix{a & b\\c & d}\rightarrow  \pmatrix{0 & 1\\1 & 0} \pmatrix{a & b\\c & d} - \pmatrix{a & c\\b & d} \pmatrix{0 & 1\\1 & 0}= \pmatrix{c & d\\a & b}-\pmatrix{c & a\\d & b}=\pmatrix{0 & d-a\\a-d & 0}$$
Or:
$g \pmatrix{a\\b\\c\\d}=\pmatrix{0\\d-a\\a-d\\0}$
So for g to have an eigenvalue, $\lambda$, we would need
$g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$
for some vector, $\pmatrix{a\\b\\c\\d}$, that is not the zero-vector. It's clear that this would require $a=0$ and $d=0$. However, any $\pmatrix{0\\b\\c\\0}$ is mapped into the zero-vector.
Therefore, $g \pmatrix{a\\b\\c\\d}= \lambda \pmatrix{0\\d-a\\a-d\\0}$  is satisfied by any vector $\pmatrix{0\\b\\c\\0}$ with eigenvalue $\lambda=0$, with two eigenvectors, $\pmatrix{0\\1\\0\\0}$ and $\pmatrix{0\\0\\1\\0}$, meaning it has a two dimensional eigenspace.
Is my thinking here correct? Am I way off? I'd appreciate any comments.
Note: I made an edit to my question, and I originally said it had no eigenvalue.
 A: Expanding my comment, it is clearer to write out what we want explicitly. We desire $\lambda$ and $A=\begin{bmatrix} a&b\\c&d \end{bmatrix}\neq 0$ such that the following holds.
$$g(\begin{bmatrix} a&b\\c&d \end{bmatrix}) = \begin{bmatrix} 0&d-a \\ a-d&0 \end{bmatrix}=\lambda\begin{bmatrix} a&b\\c&d \end{bmatrix}.$$
We separate for clarity the cases $\lambda=0$ and $\lambda\neq 0$.
For $\lambda=0$, we see that $d=a$ is the only restriction, so that $b,c$ can be chosen freely. Hence, the eigenspace corresponding to 0 is three dimensional. It has basis
$$\begin{bmatrix} 0&1\\0&0 \end{bmatrix},\begin{bmatrix} 0&0\\1&0 \end{bmatrix},\begin{bmatrix} 1&0\\0&1 \end{bmatrix}.$$
For $\lambda\neq 0$, we see that $a=d=0$ since $\lambda a = \lambda d=0$.  This forces $b=c=0$ since $\lambda c = a-d = 0 = d-a = \lambda b$. Thus there are no eigenvalues that are nonzero.
A: You are correct in your proof that $\lambda$ must be $0$.
However, you have then overlooked the eigenvector $\pmatrix{1\\0\\0\\1}$.
A: Your computations are fine (well, see the comment that you got). However, after having got the linear map$$g(a,b,c,d)=(0,d-a,a-d,0),$$you can stay that its matrix with respect to the standard basis is$$M=\begin{bmatrix}0&0&0&0\\-1&0&0&1\\1&0&0&-1\\0&0&0&0\end{bmatrix},$$whose characteristic polynomial is $\lambda^4$. So, the only eigenvalue of $M$ is $0$. Now, all that remains to be done is to solve the system$$\left\{\begin{array}{l}0=0\\-a+d=0\\a-d=0\\0=0.\end{array}\right.$$
