TITLE: In a chain of Linear Transformations, Range and Null-space Proof.
QUESTION [1 upvotes]: So to start, it is at that point in my class where the somewhat disparate definitions and theorems start to come together. The problem is that my professor isn't very good at relating them all together in a way that makes sense to me.
So here is the question:
Suppose that $Z$, $V$, $W$, $X$ are all vector spaces over the same field, and that $S : Z \to V$, $T : V \to W$, $U : W \to X$ are all linear transformations, and assume that $S$ and $U$ are invertible. Show that $R ( T \circ S ) = R ( T )$ and $N ( U \circ T ) = N ( T )$.
When I look at a simple handwritten graphing of the chain of transofrms from the question, it seems obvious to me that the range of the composite of the second and first chain would just be the range of the second chain, and that the null-space of the composite of the third chain and the second chain would just be the null-space of the second chain. I think maybe the compositions are what is tripping me up here. I'm having trouble putting my finger on a pattern to follow here for a proof. Also I'm sure I should be using the invertibility somewhere, but I'm not seeing the relation of that to the problem here.

REPLY [0 votes]: The chain of linear maps is:
$$ Z \xrightarrow{S} V  \xrightarrow{T} W  \xrightarrow{U} X  $$
There are four countainmnents to show:

$  \operatorname{range}(T \circ S) \subseteq \operatorname{range}(T) $
$  \operatorname{range}(T \circ S) \supseteq \operatorname{range}(T) $
$  \operatorname{null}(U \circ T) \subseteq \operatorname{null}(T) $
$  \operatorname{null}(U \circ T) \supseteq \operatorname{null}(T) $

Now $A \subseteq B $ means if $a \in A $ then $a \in B $.
Containments 1 and 4 are true for any linear maps. Containments 2 and 3 use $ S $ is surjective and $U $ injective respectively. 
As a guide to structuring a proof here's my proof of 2.
Choose an arbitrary element of $  \operatorname{range}(T) $, say $ T (v) \in  \operatorname{range}(T) $ where $ v \in V $. 
Since $S $ is surjective there exists $ z \in Z$ (namely $S^{-1}(v) ) $ such that $S (z) = v $.
Hence  $ T (v) = T (S (z)) \in  \operatorname{range}(T \circ S)$
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