0
$\begingroup$

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 :)

$\endgroup$
0
$\begingroup$

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

Guide:

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 $v$.

$\endgroup$
  • $\begingroup$ Thanks for the help! :) $\endgroup$ – esheesheje Nov 12 '17 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.