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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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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6750/discussion-between-dmonopoly-and-david-mitra)2012-12-16

2 Answers 2

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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!