Show that a non-zero nilpotent matrix is not diagonalizable.

I don't know how to prove that the eigenvalues for a nilpotent matrix are 0. I need a proof and not example. Thanks :)


1 Answer 1


As mentioned, by you, it suffices to prove that the eigenvalue is $0$.


Let $A$ be a nilpotent matrix, so there exists a $k$ such that $A^k = 0$.

Let $v$ be an eigenvector of $A$, so $Av=\lambda v$ and $v\neq 0$.

Hence $A^2v=\lambda^2 v$, $A^3v=\lambda^3 v$ and so on.

Evaluate $A^kv$ in terms of $\lambda$ and $v$.

  • $\begingroup$ Thanks for the help! :) $\endgroup$
    – esheesheje
    Nov 12, 2017 at 2:56

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