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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!

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    $\begingroup$ Are you aware of Jordan Canonical form? $\endgroup$ – Omnomnomnom Feb 20 '15 at 14:45
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    $\begingroup$ Jordan canonical form solves the problem, but it's overkill here. $\endgroup$ – Matt Samuel Feb 20 '15 at 14:50
  • $\begingroup$ Hint: Schur's decomposition. $\endgroup$ – Git Gud Feb 20 '15 at 14:57
  • $\begingroup$ 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. $\endgroup$ – user187039 Feb 20 '15 at 15:05
  • $\begingroup$ @user187039 Well, can you solve the problem for $J$? $\endgroup$ – Git Gud Feb 20 '15 at 15:06
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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.

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