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If a linear transformation $T:V\rightarrow V$ is injective then is it onto?

Since $dim(V)=dim(Ker(T))+dim(Im(T))$ we get $dim(V)=dim(Im(T))$ and since $Im(T)\subseteq V$ then $Im(T)=V$, therefore $T$ is onto.

Is it ok? That would mean that every linear transformation that's injective is also onto, seems weird..

McLovin
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1 Answers1

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Exactly, that's right. More generally, if $\dim V = \dim W < \infty$, and $T:V \to W$ is linear then the following statements are all equivalent:

  • $T$ is injective (one-to-one).
  • $T$ is surjective (onto).
  • $T$ is bijective (one-to-one and onto), and hence a linear isomorphism.

This is indeed a very nice result, but note that this is very specific to linear transformations. If $T:V \to W$ was not linear, then injectivity, surjectivity have nothing to do with each other.

peek-a-boo
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    Of course you made this very clear in your answer, but I want to doubly emphasize the importance of the assumption that $\dim V = \dim W$: without this assumption these three statements are not equivalent, and it's easy to forget this when you're just starting to learn linear algebra! – diracdeltafunk May 22 '20 at 02:11