# Can any linear map, $\phi$, can be expressed a as $\phi = S + N - N^T,$ where $S$ is a self-adjoint and $N$ is a nilpotent map

Inspired from this and this , I claim that

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.

## Question:

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.

• @user1551 please see my edit. – onurcanbektas Apr 13 '18 at 10:44

$A=1/2(A+A^T)+1/2(A-A^T)$.
And skew-symmetric part has $0$'s on diagonal. Then two parts of skew-symmetric part are certainly nilpotent. So for real matrices decomposition $A=\text{sym}(A)+N-N^T$ is always possible.