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Given two vector spaces A and B that have an equal number of dimensions, is any linear map from A to B that is surjective also guaranteed to be injective?

user1340033
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    When you say $A$ and $B$ have equal dimension, do you mean that $A$ and $B$ have equal finite dimension? If so, are you aware of the Rank-Nullity theorem? – Alex Wertheim Aug 16 '16 at 22:26
  • Yes I did mean finite. I'm just learning about linear maps so I haven't heard of most theorems in this space. I'll read up on the Rank-Nullity theorem. – user1340033 Aug 17 '16 at 00:57
  • Just as a comment, since noone explicited mentioned an example in the non-finite case: the left-shift (or "forgets-first-coordinate" function) on $\mathbb{R}^{\mathbb{N}}$ is obviously surjective and obviously not injective. – Aloizio Macedo Aug 26 '16 at 09:00

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If $A$ and $B$ have the same (finite) dimension, a linear map is surjective if, and only if, is injective. If $\phi:A\longrightarrow B$ is a linear map, then $$dim(A)=dim(ker(\phi))+dim(Im(\phi)).$$ Therefore, $\phi$ is surjective if, and only if, $\phi$ is injective.

Rafael
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Yes. It follows from Rank-Nullity Theorem.

Renan R.
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Yes. Hint 1: Being of finite dimension is a necessary hypothesis. Hint 2: Regular functions between sets of the same size have the property injective iff. surjective.

Use the finite basis!

operatorerror
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