# Thread: [SOLVED] Linear Algebra, Nilpotent

1. ## [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.

2. 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?

3. 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

4. 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.

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

6. 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.