# Every $n\times n$ matrix is the sum of a diagonalizable matrix and a nilpotent matrix.

I would like to prove that every $n\times n$ matrix is the sum of a diagonalizable matrix and a nilpotent matrix. How is this possible? I'm not sure where to begin really- I know that a nilpotent matrix is one of which some power is the zero matrix. I also know that a matrix A can be written as $AP=PJ$ with $P$ invertible and $J$ of Jordan form. I have proven that any strictly upper triangular matrix is nilpotent, so $J$ can be written as $D+N$, with D diagonal and $N$ nilpotent, but how can I change this for A? Thank you!

• Are you aware of Jordan Canonical form? – Omnomnomnom Feb 20 '15 at 14:45
• Jordan canonical form solves the problem, but it's overkill here. – Matt Samuel Feb 20 '15 at 14:50
• Hint: Schur's decomposition. – Git Gud Feb 20 '15 at 14:57
• Yes I am aware the Jordan form, I know that given a matrix A, AP=PJ where P is invertible and J is in Jordan form. – user187039 Feb 20 '15 at 15:05
• @user187039 Well, can you solve the problem for $J$? – Git Gud Feb 20 '15 at 15:06

You have $A = PJP^{-1}$ where $J$ is in Jordan form. Write $J = D + N$ where $D$ is the diagonal and $N$ is the rest, which is strictly upper triangular and thus nilpotent. Then $A = PDP^{-1} + PNP^{-1}$. The former is clearly diagonalizable, while the latter is nilpotent; just note that $(PNP^{-1})(PNP^{-1}) = PN(P^{-1}P)NP^{-1} = PN^2P^{-1}$ and so on.