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Say V is a finite dimensional vector space, and $T_1: V \to\ V$ and $T_2: V \to\ V$ are linear transformations. Assume that $T_1$ is one to one and $T_2$ is onto.

I am uncertain if my proof that $T_1T_2$ is an isomorphism is valid.

If you don't want to look at the picture - Is it necessary to show that $T_1$ and $T_2$ are each isomorphisms themselves by Rank-Nullity?

Here is an image of my work: Math proof

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(In case anyone is wondering, this is optional, extra review for a class. The proof is definitely not formal.) –  dmonopoly Dec 15 '12 at 23:42
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The kernal of $T_1$ is $\{{\bf 0}\}$. So if $T_1(T_2(v))=\bf0$, you can deduce $T_2(v)=\bf 0$. But how do you surmise $v=\bf 0$? I think the best approach would be to show both $T_1$ and $T_2$ are isomorphisms. Then show the composition is. –  David Mitra Dec 15 '12 at 23:52
    
I see how $T_2(v) = 0$ if the composition = 0, but I don't follow the "then show the composition is [an isomorphism]"... –  dmonopoly Dec 16 '12 at 0:29
    
Sorry, the last two sentences in my previous comment were a different topic... I meant to suggest that you just redo the whole proof (what you have in the image doesn't seem complete to me). First show that both $T_1$ and $T_2$ are isomorphisms (you just need to show $T_1$ is onto and $T_2$ is one-to-one). Then show $T_1T_2$ is an isomorphism. –  David Mitra Dec 16 '12 at 0:39
    
I understand how to show $T_1$ is onto and $T_2$ is one-to-one, but afterward, I don't see how I would use that to show $T_1T_2$ is an isomorphism (that's where I get stuck). –  dmonopoly Dec 16 '12 at 0:42

2 Answers 2

I think it is easier to show the following easy claim which follows at once from the dimension (or rank-nullity, if we talk matrices language) theorem:

Claim: An operator $\,T:V\to V\,$ , with $\,\dim V<\infty\,$ is bijective iff it is injective iff it is surjective.

With the above, it follows at once that both $\,T_1,T_2\,$ are bijections (isomorphisms of vector spaces)...

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Thanks to a discussion with @David_Mitra, I have an answer that I understand. In short, my picture answer is not sufficient, and I do need to show that $T_1$ and $T_2$ are each isomorphisms.

More specifically, we just need to make sure that $T_1$ and $T_2$ are each one-to-one:

  1. By Rank-Nullity, we can show that $T_2$ is not only onto, but also 1-1.
  2. Observe $T_2(v) = 0$ if $T_1(T_2(v))=0$ because $Ker(T_1) = {0} $ (because $T_1$ is 1-1)
  3. Observe $v = 0$ if $T_2(v)=0$ because $Ker(T_2) = {0} $ (because $T_2$ is 1-1)
  4. Therefore, $Ker(T_1T_2)=Ker(T_1(T_2(v)))=0$ - so $T_1T_2$ is 1-1
  5. By Rank-Nullity, we can show that $T_1T_2$ is onto
  6. $T_1T_2$ is an isomorphism!
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