# Nilpotent matrix

1. Jul 19, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
Show that strictly upper triangular $n\times n$ matrices are nilpotent.

2. Relevant equations

3. The attempt at a solution

Let $f$ be the endomorphism represented by the strict upper triangular matrix $M$ in basis ${\cal B} = (e_1,...,e_n)$.
We have that $f(e_k) \in \text{span}(e_1,...,e_{k-1})$, $(f\circ f)(e_k)\in f(\text{span}(e_1,...,e_{k-1}))= \text{span}(f(e_1),...,f(e_{k-1})) \subset \text{span}(e_1,...,e_{k-2})$... Repeating this process, we are sure that $f^{(k)}(e_k) = 0$. So $\ell\ge n \Rightarrow M^\ell = 0$, right?

2. Jul 19, 2015

### Zondrina

So you want to show $M_{ij}^k = 0$ for some positive integer $k$.

If I'm reading your post correctly, you're saying $f(e_k) = \text{Some strictly upper triangular matrix M}$ for any basis vector in $\cal B$.

I think it looks okay, but the notation is a little confusing. When you write $f^{(k)}(e_k) = 0$, some people may get confused, and so I think it is better to write it as:

$${(f(e_k))}^k = 0$$

To signify you want the $\text{k}^{th}$ power of the morphism of the ${e_k}^{th}$ basis vector.

If ${(f(e_k))}^k = M_{ij}^k = 0$ for some positive integer $k$, then you can go as far as to say $M_{ij}^{\ell} = 0, \forall \ell \geq k$. This is intuitive because eventually with so many powers of the matrix, there will be enough zeroes to multiply and produce the zero matrix. Then you can assume every matrix power afterwards is the zero matrix.

Last edited: Jul 19, 2015
3. Jul 19, 2015

### Zondrina

There is an alternate way to prove this:

Suppose $A \in M_{n \times n}( \mathbb{F} )$ is a strictly upper triangular matrix. Then $A$ has characteristic polynomial $x^n$. Using the the fact:

$$p( \lambda ) = \text{det}(A - \lambda I)$$

You can deduce $A^n = 0$.

4. Jul 19, 2015

### geoffrey159

Hello,

Yes

$f(e_k)$ is the k-th column of matrix $M$, which is strictly upper triangular

By $f^{(k)}$, I meant the k-th composition by $f$ ($f\circ ... \circ f$ k times), not the k-th power.

$f^{(k)}(e_k) = 0$ means the k-th column is zero after k multiplications of M by itself, not that the whole matrix is zero, right ?

I don't understand, could you elaborate please?

5. Jul 19, 2015

### vela

Staff Emeritus
Look up the Cayley-Hamilton theorem.

6. Jul 19, 2015

### Zondrina

I was quite confused by the notation when I first looked at it.

If you want help with the alternate proof, show some of your thoughts.

7. Jul 20, 2015

### geoffrey159

Sorry, I was very lazy yesterday.
Your idea is the most simple math proof ever ! My thought on this is (Wikipedia's thought ) is that the caracteristic polynomial is zero in $A$, so that $0 = p(A) = (-1)^n A^n \iff A^n = 0$.
What a nice idea !