Any linear map, $\phi$ on a real vector space $E$, can be written as a
$$\phi = S + N - N^T,$$ where $S$ is a self-adjoint and $N$ is a nilpotent map ($N^T$ denotes the adjoint of $N$, and its matrix is the transpose of the matrix of $N$)
Observe that, the map $$\psi = N - N^T$$ is a skew map. Moreover, given a skew, map we can always find a nilpotent map $N$ s.t the above equality holds.
In fact, we we take the entries of the matrix of $\psi$ above diagonal and form a new matrix with those entries, we get a nilpotent matrix, and the above equality holds.
As it has been proven that the linked question, we have
$$\phi = S + \psi,$$ hence we have $$\phi = S + N - N^T$$ QED.
Is there anything wrong with this argument ? I mean I have never seen such a theorem, anywhere, and I want to make sure that the relation that I found is correct.