Proof of Nilpotent Matrix: Strictly Upper Triangular Matrices

[geoffrey159]

Homework Statement

Show that strictly upper triangular $n\times n$ matrices are nilpotent.

Homework Equations

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?

 

[STEMucator]

Homework Statement

Show that strictly upper triangular $n\times n$ matrices are nilpotent.

Homework Equations

The Attempt at a Solution

Let $f$ be the endomorphism represented by the strictly 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?

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.

 

[STEMucator]

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 $A^n = 0$.

 

[geoffrey159]

Hello,

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

Yes

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 $\cal B$.

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

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.

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

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.

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

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 $A^n = 0$.

I don't understand, could you elaborate please?

 

[vela]

I don't understand, could you elaborate please?

Look up the Cayley-Hamilton theorem.

 

[STEMucator]

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

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.

 

[geoffrey159]

Look up the Cayley-Hamilton theorem.

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.

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 !