So here is the questions I am trying to do
How do I prove this I am really struggling! I know diagonalised matrix may have complex values for even real A, so i am not sure this questions is correct. I guess there has to be something with L + N , because left and right should be real ... but complex conjugate may cancel out to give u real values
The idea is as follows: note that $$ \bar L + \bar N = \bar A = A. $$ Moreover, $\bar L$ is diagonalizable, $\bar N$ is nilpotent, and $\bar L \bar N = \bar N \bar L$. Because the Jordan-Chevalley decomposition of an operator is unique, it must hold that $\bar L$ is equal to the "diagonal part" of $A$, which is to say that $\bar L = L$, and $\bar N$ is equal to the "nilpotent part" of $A$, which is to say that $\bar N = N$.
For a more direct proof, note that there exist polynomials $p,q$ such that $L = p(A)$ and $N = q(A)$. It follows that because $\bar N, \bar L$ commute with $\bar A = A$, they compute with both $N$ and $L$. Now, we note that $$ L + N = \bar L + \bar N \implies L - \bar L = N - \bar N. $$ Because $L, \bar L$ commute and are diagonalizable, $L - \bar L$ is diagonalizable. Because $N,\bar N$ commute and are nilpotent, $N - \bar N$ is nilpotent. The only operator that is diagonalizable and nilpotent is the zero operator, so $N - \bar N = L - \bar L = 0$.