# Nilpotent

• Apr 18th 2010, 09:14 AM
dwsmith
Nilpotent
An nxn matrix is said to be nilpotent if $A^k=0$ for some positive $\mathbb{Z}$ k. Show that all eigenvalues of a nilpotent matrix are 0.

I have proved by math induction that, for $m\geq1$, $\lambda^m$ is an eigenvalue of $A^m$.

I don't know if that should help.
• Apr 18th 2010, 09:22 AM
HallsofIvy
Quote:

Originally Posted by dwsmith
An nxn matrix is said to be nilpotent if $A^k=0$ for some positive $\mathbb{Z}$ k. Show that all eigenvalues of a nilpotent matrix are 0.

I have proved by math induction that, for $m\geq1$, $\lambda^m$ is an eigenvalue of $A^m$.

I don't know if that should help.

Yes, that certainly does help!(Clapping) I presume that in doing that you also showed that if v is an eigenvector of A corresponding to eigenvalue $\lambda$ then v is also an eigenvector of $A^m$ corresponding to eigenvalue $\lambda^m$.

In particular, if $\lambda$ an eigenvalue of A, then $\lambda^k$ is an eigenvalue of $A^k$- that is, for some non-zero vector v, $A^k v= \lambda^k v$. But $A^kv= 0$ for any vector so we have $\lambda^k v= 0$, with v non-zero.
• Apr 18th 2010, 09:29 AM
dwsmith
Quote:

Originally Posted by HallsofIvy
Yes, that certainly does help!(Clapping) I presume that in doing that you also showed that if v is an eigenvector of A corresponding to eigenvalue $\lambda$ then v is also an eigenvector of $A^m$ corresponding to eigenvalue $\lambda^m$.

In particular, if $\lambda$ an eigenvalue of A, then $\lambda^k$ is an eigenvalue of $A^k$- that is, for some non-zero vector v, $A^k v= \lambda^k v$. But $A^kv= 0$ for any vector so we have $\lambda^k v= 0$, with v non-zero.

So that is all it is then?
• Apr 18th 2010, 09:59 AM
HallsofIvy
Well, what do you conclude from $\lambda^k v= 0$?
• Apr 18th 2010, 10:11 AM
dwsmith
I should probably conclude lambda is zero.