Understanding proof for theorem about dimension of kernel

[Incand]

So the theorem says:
Suppose that $U$ and $V$ are finite dimensional vector spaces, and that $T:U\to V$, $S: V \to W$. Then
$\text{dim Ker }ST \le \text{dim Ker }S + \text{dim Ker }T$.

Proof:
Set $U_0 = \text{Ker }ST$ and $V_0 = \text{Ker }S$. $U_0$ and $V_0$ are subspaces of $U$ and $V$. Since $ST=0$ on $U_0$ we have $T(U_0) \subset V_0$. We can then consider $T$ and $S$ as mappings defined on $U_0$ respectively $V_0$. Since $\text{Ran }T$ (Range, column space, image etc.) is a subspace to $V_0$ we have $\text{dim Ran }T \le \text{dim }V_0 = \text{dim Ker }S$. But according to the Rank-nullity theorem, $\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0$.
Hence $\text{dim Ker }ST = \text{dim }U_0 = \text{dim Ker T} + \text{dim Ran }T \le \text{dim Ker }T + \text{dim Ker }S$.

(I translated this so it's possible worse worded than originally.)

I can't really follow the steps in this proof, no matter how many times I look through it I don't follow all the steps. Starting it with
$U_0$ and $V_0$ are subspaces of $U$ and $V$
The word "and"" confuse me here. It's supposed to mean that $U_0$ to $U$ and $V_0$ to $V$ and nothing else right?

We can then consider $T$ and $S$ as mappings defined on $U_0$ respectively $V_0$.
What does this mean? That for our purposes we can consider $T:U_0 \to V$ and $S:V_0 \to W$? Why?

Since $\text{Ran }T$ is a subspace to $V_0$
I understand this to be true if we indeed consider $T:U_0 \to V$ since by definition $T(U_0)$ maps into $V_0$ but not for $U$ in general.
And then $\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0$. I don't get how we're allowed to only consider $U_0$ instead of $U$.

 

[Erland]

I can't really follow the steps in this proof, no matter how many times I look through it I don't follow all the steps. Starting it with
$U_0$ and $V_0$ are subspaces of $U$ and $V$
The word "and"" confuse me here. It's supposed to mean that $U_0$ to $U$ and $V_0$ to $V$ and nothing else right?

Right. But the wording in is unclear in the passage. The word "respectively" should have been added.

We can then consider $T$ and $S$ as mappings defined on $U_0$ respectively $V_0$.
What does this mean? That for our purposes we can consider $T:U_0 \to V$ and $S:V_0 \to W$? Why?

The author actually takes the restriction of $S$ to $V_0$ and the restriction of $T$ to $U_0$.

Since $\text{Ran }T$ is a subspace to $V_0$
I understand this to be true if we indeed consider $T:U_0 \to V$ since by definition $T(U_0)$ maps into $V_0$ but not for $U$ in general.
And then $\text{dim Ker }T + \text{dim Ran }T = \text{dim }U_0$. I don't get how we're allowed to only consider $U_0$ instead of $U$.

The author is working with the restrictions, which have the domains $U_0$ and $V_0$, respectively, and then there is no problem, right?

 

[Incand]

The author actually takes the restriction of $S$ to $V_0$ and the restriction of $T$ to $U_0$.

I don't understand this, perhaps you could write out the transformation if what I wrote was wrong. $S$ is a mapping from $V$ to $W$ and $T$ a mapping from $U$ to $V$ then $ST: U_0 \to V_0 \to 0$ in my mind. $S$ can't map into $V_0$ since $V_0$ may not be a subspace of $W$.

 

[Erland]

You got this essentially correct, just that $V_0$ is a subspace of $V$, not of $W$. And kernels are of course subspaces of the spaces they lie in.

Edit: (The restriction of) $S$ maps $V_0$ into $0$. $ST$ means that $T$ is applied first, then $S$.

 

[Incand]

You got this essentially correct, just that $V_0$ is a subspace of $V$, not of $W$. And kernels are of course subspaces of the spaces they lie in.

Edit: (The restriction of) $S$ maps $V_0$ into $0$. $ST$ means that $T$ is applied first, then $S$.

Alright I understand that part. I think I finally also understand why it's equivalent to make the proof for the "new" $T$. Since the theorem involve only kernel of the operators (and not every other vector) it's equivalent to prove the theorem for an operator mapping from only the kernel's. That the steps in the proof ain't actually true for a general operator $S$ and $T$ doesn't matter since we already know it's equivalent to show the theorem for the new $T$ and $S$.

It seems straightforward now, thanks! The problem I had was pretty much I felt that the most crucial step that it's equivalent to prove it for the "new" operators wasn't written out. I think if I written the proof myself would've emphasized that since for me it's was really the key step in the proof.

 

[Erland]

Yes, I think the proof would be more pedagogical if one introduces and uses the restriction $T_0$ of $T$ to $U_0$ with codomain $V_0$, and the restriction $S_0$ of $S$ to $V_0$ with codomain $0$.

 

[mathwonk]

If a vector x goes to zero under a composition ST then T(x) must go to zero under S, so ker(ST) is contained in the pullback of the kernel of S under T. But pulling back a subspace increases the dimension by at most that of the kernel, so the pullback of ker(S) has dimension ≤ kerS + kerT. Since this pullback contains ker(ST) we have the inequality. I did this in my head, but I think it is probably right. I was trying to think of an argument with some geometric motivation.

 

[Incand]

That's some nice reasoning mathwonk. I be sure to remember to take a closer look at it in the future when I have a bit better math knowledge.