Show that (i) $\dim\ker T=\dim\ker ST$ (ii) $\dim\operatorname{im}T=\dim\operatorname{im}ST$ 
Let $T: V \rightarrow W$ be a linear mapping and $S: W \rightarrow U$ be an isomorphism (all finite dimensional vector spaces). Show that
(i) $\dim\ker T=\dim\ker ST$
(ii) $\dim\operatorname{im}T=\dim\operatorname{im} ST$

Attempt;
Clearly, $T: V \rightarrow W$; $S: W \rightarrow U$ then $ST:V\rightarrow U$
Also $\dim\ker T+\dim\operatorname{im} T=n=\dim V$
$\dim\operatorname{im}S=m=\dim W$ as $\dim\ker S=0$, for $S$ is isomorphism
and $\dim\ker ST+\dim\operatorname{im}ST=n=\dim V$
what to do next?
 A: $\newcommand{\im}{\operatorname{im}}$I think the easiest way to do this would be to directly show $\ker T=\ker ST$ (which I encourage you do to on your own; of course $\ker T\subseteq\ker ST$ is trivial so you just need to do the reverse). With this in mind, using rank-nullity we have
$$\dim V=\dim\ker T+\dim\im T$$
and
$$\dim V=\dim\ker ST+\dim\im ST$$
However, since you know $\dim\ker T=\dim\ker ST$, you can set the two equations equal and recover $\dim\im T=\dim\im ST$.
A: $Tv = 0 \implies STv =0 \implies \ker T \subset \ker ST$. 
Since $S$ is an isomorphism, it is injective so that $\ker S = \{0\}$. 
$x \in \ker ST \implies STx = 0 \implies Tx \in \ker S = \{0 \} \implies Tx = 0 \implies x \in \ker T$.
For the second part use the rank-nullity theorem. 
A: I'd suggest thinking about this a bit less mechanically -- instead, think about what these things actually represent.
An isomorphism like $S$ identifies two spaces as being identical, after renaming (with $S$ providing the renaming).
Further, this is fundamentally a question about function composition; so, let's think about how things work for function composition in general.
First: what is the kernel of $ST$?  It is the set of all $v\in V$ such that $T(v)\in\ker S$.  But, $S$ is an isomorphism; so, $\ker S$ is precisely the $0$-vector in $W$.  So, $\ker ST$ is exactly the set of $v\in V$ such that $T(v)=\vec{0}_W$... which is precisely $\ker T$.
For (ii), you can play similar tricks; but, you can also realize that $\text{im} ST=S(\text{im} T)$, and $S$ is an isomorphism.
A: For the second part consider the following argument.
$Im(ST)=\{(ST)(x):x\in V\}\;=\{S(T(x)):x \in V\}\;=Im(S_{|Im(T)})$ 
where, $S_{|Im(T)}$ is a restriction map from $Im(T)$ onto its image. 
Since $S$ is an isomorphism $\implies$ $S_{|Im(T)}$ is too.
Thus, $dim(Im(T))\;=dim(Im(S_{|Im(T)}))\;=dim(Im(ST))$.
