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?
Best Answer
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. $v=Tu$.