linear endomorphism prove $T^n = 0$ where n = dimV

Question

Let $T : V \rightarrow V$ be a linear endomorphism on a vector space $V$ of dimension $n$.

Suppose that $T^k = 0$ for some positive integer $k$.

Show that $T^n = 0$.

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I have tried it and found that there is a example.

Let $V = \{1 ,x ,x^2\}$ , $T:=f'(x)$

and actually $T^{dim_FV} = 0 $

But still no ideas how to prove it.

Could anyone give me hints?

I just learn vector space (chap1) and linear map (chap2) without matrix or determinant in book, and the problem arrange in linear map (chap 2).

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I try to search the solution, and find the link

http://www.math.ualberta.ca/~xichen/math32515w/hw5sol.pdf

Page 5 , Problem 5

but I can't understand the inequalities

Why $\geq$ can be $>$ ?

Answer 1

This proof is more or less analoguous to the one you showed in your question, with the inequalities and the ranks, except from the perspective of subspaces. Take the (infinite) chain of vector spaces $$ V\supseteq TV\supseteq T^2V\supseteq\cdots\supseteq T^nV\supseteq\cdots $$ where $T^nV = \{T^nv\mid v\in V\}$ is the image of $T^n$ (swapping out $T^iV$ with $\operatorname{rank}(T^i)$ and $\supseteq$ with $\geq$ gives your chain of inequalities). Some (most) of these inclusions are equalities, and for every single one that is not an equality, the dimension goes down by at least $1$ from $T^iV$ to $T^{i+1}V$ (if one vector space is contained in some other vector space, and their dimensions are finite and equal, then the vector spaces are equal).

Also, once we hit an equality, all the rest are necessarily equalities. We see this because if $T^iV = T^{i+1}V$, then $$T^{i+2}V = T(T^{i+1}V) = T(T^iV) = T^{i+1}V$$ and similarily, $T^{i+3}V = T^{i+2}V$, and so on. (Note that $T^iV = T^{i+1}V$ does not mean $T^i = T^{i+1}$. For instance, if $V$ is the plane and $T$ is a rotation, then $TV = T^2V$, but $T\neq T^2$.)

So once we hit an equality it's all equalities. And if an inclusion is not an inequality, the dimension goes down by at least $1$. Since the chain starts with $V$ at dimension $n$, that necessarily means that the absolute longest we can possibly keep the chain going without hitting equalities is $n$ steps, until $T^nV$. We therefore must have $$ T^nV = T^{n+1}V = T^{n+2}V = \cdots $$ Everything I've said here is entirely general and true for any (finite-dimensional) vector space $V$ and linear operator $T:V\to V$. This also means that if any of the vector spaces in the chain is the zero space, like in your case, where we're told that $T^k = 0$, then $T^nV$ must be the zero space.

Answer 2

Hint:

What can you say on the minimal polynomial of $T$?

Answer 3

Let $P(x)$ be the minimal polynomial of $T$. It follows from the Cayley-Hamilton that $\deg P(x)\leqslant n$. So, if $T^k=0$, $P(x)\mid x^k$. It follows that $P(x)$ is a power of $x$ and, since $\deg P(x)=d$ for some $d\leqslant n$, $P(x)\mid x^n$. Therefore, $T^n=0$.

Answer 4

If $T^m=0$ and $T^{m-1} \ne 0$, then there is $v \in V$ such that $T^{m-1} v \ne 0$. Then $v, Tv, T^2 v, \dots,T^{m-1} v$ are linearly independent. Therefore, $m \le n$.

Answer 5

Let $m$ be the minimal positive integer such that $T^m = 0$. Since $\text{Im}\ T^{i+1} = T^i(\text{Im}\ T) \subseteq \text{Im}\ T^i$ we have an ascending chain of linear subspaces:

$$\{0\} = \text{Im}\ T^{m} \subseteq \text{Im}\ T^{m-1}\subseteq\ldots\subseteq \text{Im}\ T\subseteq \text{Im}\ T^{0} = V$$

In particular, if $\dim \text{Im}\ T^{i+1} = \dim \text{Im}\ T^{i}$, then we must already have $\text{Im}\ T^{i+1} = \text{Im}\ T^{i}$. But if this ever happens then:

$$ \{0\} = \text{Im}\ T^{m} = T^{m-{i+1}}(\text{Im}\ T^{i+1}) = T^{m-(i+1)}(\text{Im}\ T^{i}) = \text{Im}\ T^{m-1}$$

Contradicting the minimality of $m$. Therefore $\dim\text{Im}\ T^{i+1}< \dim \text{Im}\ T^{i}$ for all $i$. Since $\dim V =n$ this implies $m\le n$