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Sorry if this is somewhat a duplicate. The answers I see deal with functions in general rather than linear maps.

Let $T$ be a linear map from $U$ to $V$.
I understand that by definition a linear map is injective if every element in the range gets mapped there by a unique vector from the domain. This is easy to show by choosing two vectors $u$ and $v$ in $U$, and showing that if $T(u)=T(v)$, then $u=v$.

But for a surjective linear map, it does not seem like there is something simple like this we can do? We have to show that range$(T)=V$. How is this done?

EDIT: As a concrete example, suppose we have $T\in L(F^\infty \rightarrow F^\infty)$ defined by $T(x_1,x_2,x_3,\dots) = (x_2,x_3, \dots)$. How can we show this is surjective? Is it enough to:

Suppose $w\in W$, where $w=(w_1, w_2, \dots)$. Then let $u=(a, w_1, w_2)$ for some $a\in F$. And that's all we need?

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When you say, "I understand that by definition a linear map is surjective", the word "surjective" should be replaced by "injective". – Zev Chonoles Apr 10 '11 at 0:59
Thanks I fixed that. – fdart17 Apr 10 '11 at 1:03
I have the same feeling as yours: often, proving that a linear map is injective is easier than proving it is surjective. Of course we are talking about infinite-dimensional vector spaces, otherwise the two are equivalent. – Giuseppe Negro Apr 10 '11 at 1:12
Oh and regards your edit: OK, you have proven that $T$ is surjective. – Giuseppe Negro Apr 10 '11 at 1:13
up vote 5 down vote accepted

If you're dealing with finite-dimensional spaces, the key is relation:

$$\tag{1} \dim U = \dim \mbox{im} T +\dim \ker T \; .$$

If you know how to evaluate $\dim \ker T$, then you can easily compare $\dim V$ with $\dim \mbox{im} T=\dim U-\dim \ker T$ and your map will be surjective iff $\dim V =\dim \mbox{im} T$.

I want to remark also a straightforward consequence of (1). If $U$ and $V$ are both finite-dimensional and they both have the same dimension, then there is equivalence between being injective and being surjective for linear maps, i.e. a linear map $T:U\to V$ is injective iff it is surjective.

On the other hand, AFAIK, when you deal with infinite-dimensional spaces surjectivity proofs cannot be shortened by using tricks: in general, one has to show that for each $v\in V$ there exists $u\in U$ s.t. $v=Tu$.

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$im$ means range correct? – fdart17 Apr 10 '11 at 4:35
@Cplayer: Correct. – Pacciu Apr 10 '11 at 9:53
(@TheChaz: From time to time I like to read some question and some (genuinely brilliant) answers... But it seems I can't stand the Q/A format of Math.SE. I don't want to be rude, but why you're asking?) – Pacciu Oct 25 '11 at 22:10
@The Chaz: I see. I'm much more active on, an Italian Math forum (I have +10K posts there, under a different nickname, and I'm one of the moderators). That forum also has a section called English Corner, where users post exercises in order to practice writing in English... Wanna join us? ;-) – Pacciu Dec 9 '11 at 2:35
All I need is another math forum... But hey, I'm inebriated; why not?!? – The Chaz 2.0 Dec 9 '11 at 4:22

For surjectivity, you need to show that for every $v \in V$ there exists a $u \in U$ such that $T(u) = v$.

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