# [SOLVED] Linear Algebra, Nilpotent

• Apr 21st 2010, 11:37 AM
JJMC89
[SOLVED] Linear Algebra, Nilpotent
Let $\displaystyle V$ be a finite dimensional vector space over $\displaystyle \mathbb C$. Let $\displaystyle T$ be a linear transformation
on $\displaystyle V$ all of whose eigenvalues are zero. Show that $\displaystyle T^{n} = 0$ for some $\displaystyle n$, i.e.
that $\displaystyle T$ is nilpotent.
• Apr 21st 2010, 12:22 PM
Defunkt
Quote:

Originally Posted by JJMC89
Let $\displaystyle V$ be a finite dimensional vector space over $\displaystyle \mathbb C$. Let $\displaystyle T$ be a linear transformation
on $\displaystyle V$ all of whose eigenvalues are zero. Show that $\displaystyle T^{n} = 0$ for some $\displaystyle n$, i.e.
that $\displaystyle T$ is nilpotent.

Do you know the Cayley-Hamilton theorem?
• Apr 21st 2010, 02:35 PM
dwsmith
Quote:

Originally Posted by JJMC89
Let $\displaystyle V$ be a finite dimensional vector space over $\displaystyle \mathbb C$. Let $\displaystyle T$ be a linear transformation
on $\displaystyle V$ all of whose eigenvalues are zero. Show that $\displaystyle T^{n} = 0$ for some $\displaystyle n$, i.e.
that $\displaystyle T$ is nilpotent.

http://www.mathhelpforum.com/math-he...nilpotent.html
• Apr 21st 2010, 03:30 PM
Defunkt
Quote:
This is not the same.

In your topic, you wanted to show that if A is nilpotent, then all of its eigenvalues are 0.

Here he wants to prove that if all of A's eigenvalues are 0, then it is nilpotent.
• Apr 21st 2010, 07:12 PM
JJMC89
Quote:

Originally Posted by Defunkt
Do you know the Cayley-Hamilton theorem?

Yes.
• Apr 22nd 2010, 04:35 AM
HallsofIvy
Well, then you have it, don't you? If all eigenvalues are 0, then the characteristic equation is $\displaystyle \lambda^n= 0$ and every matrix satisfies its own characteristic equation.