Vector space sizes

Question

$T:V \rightarrow V$ where $V$ is a finite dimensional real vector space

I can show that $\ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3) \subseteq\cdots $

Prove there exists some $k$ such that $\ker(T^k)=\ker(T^{k+j}) $

$\forall j \geq 0 $

My plan: show there exists some $k$ such that $\ker(T^k)=\ker(T^{k+1}) $ I have already shown via rank-nullity and induction that if this is true the bit we are trying to prove will also be true. But I don't want to digress with that.

In order to show there exists some $k$ such that $\ker(T^k)=\ker(T^{k+1}) $

Assume there exists no such $k$

Then $\ker(T) \subset \ker(T^2) \subset \ker(T^3) \subset \cdots $ (strict subsets)

so the size of the set $\ker(T^n)$ is strictly increasing. I was hoping to reach some kind of contradiction because $\ker(T^n)$ is a subspace of $V$ -well it's a subset which is the main thing. So it cant keep increasing forever... but then $V$ may not be a finite vector space so not really sure what to do...

Question: It makes no sense to say a vector space is finite does it?- it must be a set of infinite size- even if a vector space has a finite basis there infinitely many linear combinations you could take of the basis elements right?

Answer 1

A few things:

Yes, a vector space can be finite:
In particular, every finite-dimensional space over a finite field is finite. The integers modulo any prime gives you a finite field. The only finite vector spaces over infinite fields like $\Bbb R$ or $\Bbb C$, however, are the trivial space $\{0\}$.

Yes, a vector space can be infinite dimensional:
Two classic examples are the space of polynomials and the space of infinite sequences.

Your result will only hold for finite dimensional spaces:
Consider the function on the space of infinite sequences given by $$ (x_1,x_2,x_3,\dots) \mapsto (x_2,x_3,x_4,\dots) $$

Yes, your result holds for all finite dimensional spaces $V$:
In particular, it suffices to note that

• $\ker T^{j} \subseteq \ker T^{j+1}$
• $\ker T^j$ is always a subspace
• if $\ker T^k = \ker T^{k+1}$, then $\ker T^k = \ker T^{k+j}$ for all integers $j \geq 0$.

And of course, if $V$ is finite dimensional, $\dim \ker T^j \leq \dim V$.

Answer 2

Hint: If $V$ is finite-dimensional, all subspaces are finite dimensional, and if if a subspace $K_1$ is contained in the subspace $K_2$, and $\dim K_1=\dim K_2$, then $K_1=K_2$.