Question
Suppose $U$, $V$ and $W$ are finite-dimensional vector spaces. Let $\mathcal{L}(U,V)$ and $\mathcal{L}(V,W)$ be the vector spaces of all linear maps from $U$ into $V$ and from $V$ into $W$, respectively. Suppose $S \in \mathcal{L}(V,W)$ and $T \in \mathcal{L}(U,V)$. Then prove that
$\begin{align}
& 1. \, \text{dim null} S \circ T \le \text{dim null} S + \text{dim null} T \\
& 2. \, \text{dim range} S \circ T \le \text{min} \{ \text{dim range} S , \text{dim range} T \}
\end{align}$
My Thought
To prove the first one, I guess that writing the fundamental theorem of linear maps for $S \circ T$ may be a good start
$$\begin{align}
\text{dim null} S \circ T &= \text{dim} U – \text{dim range} S \circ T \\
&= \text{dim null} T + \text{dim range} T – \text{dim range} S \circ T \\
&\le \text{dim null} T + \text{dim} V – \text{dim range} S \circ T \\
&= \text{dim null} T + \text{dim null} S + \text{dim range} S- \text{dim range} S \circ T
\end{align}$$
So if I can prove that
$$\text{dim range} S- \text{dim range} S \circ T \le 0$$
then I am done but this does not seem to be true because it is easy to see that $\text{range} S \circ T \subseteq \text{range} S$ and hence
$\text{dim range} S \circ T \le \text{dim range} S$
. So I am stuck!
Also, I could observe that $\text{null} T \subseteq \text{null} S \circ T$ and hence
$\text{dim null} T \le \text{dim null} S \circ T$
but I don't know how to use this!
Best Answer
Fo the first question, I would argue like this: let $K = \mathrm{null}(ST) = T^{-1}(\mathrm{null}(S))$, a subspace of $U$, and let $T|_{K} : K \to V$ be the restriction of $T$ to $K$. Because
\begin{align*} \mathrm{null}(T|_{K}) \subset \mathrm{null}(T) && \text{and} &&\mathrm{ran}(T|_{K}) \subset \mathrm{null}(S) \end{align*}
we get
\begin{align*} \dim(\mathrm{null}(ST)) &= \dim(K) \\ &= \dim(\mathrm{null}(T|_{K})) + \dim(\mathrm{ran}(T|_{K})) \\ &\leq \dim(\mathrm{null}(T)) + \dim(\mathrm{null}(S)) \end{align*}
as desired.
For your second question, note what you need to do is prove (i) $\dim(\mathrm{ran}(ST))≤\dim(\mathrm{ran}(S))$ and (ii) $\dim(\mathrm{ran}(ST))≤\dim(\mathrm{ran}(T))$. I think you can probably see why (i) is true. For (ii), note that $ST$ has the same range as $S|_{\mathrm{ran}(T)}$, and that the range of a linear tranformation has equal or lesser dimension than the domain