# Dimension of Range and Null Space of Composition of Two Linear Maps

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!

## 1 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

• (+1) Thanks for the attention. :) Two questions: 1. Why $\mathrm{null}(ST) = T^{-1}(\mathrm{null}(S))$? 2. What about the case number two in the question? Would you please write something for that too. :) Commented Aug 21, 2016 at 19:08
• For your first question, it just comes down to definitions: $x \in \mathrm{null}(ST)$ iff $STx = 0$ iff $Tx \in \mathrm{null}(S)$ iff $x \in T^{-1}(\mathrm{null}(S))$. For your second question, note what you need to do is prove (i) $\dim(\mathrm{ran}(ST)) \leq \dim(\mathrm{ran}(S))$ and (ii) $\dim(\mathrm{ran}(ST)) \leq \dim(\mathrm{ran}(T))$. 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. Commented Aug 21, 2016 at 19:18
• OK the confusion here is over notation. I was writing $T^{-1}$ to indicate taking the inverse image of a set, I was not assuming $T$ is invertible. Here, it is important to note that the inverse image of a subspace is a subspace---something worth working out if this is not a familiar fact! Commented Aug 21, 2016 at 19:23
• About the inclusions, I get $\mathrm{null}(T_0) = \mathrm{null}(T)$ and $\mathrm{ran}(T_0) \subseteq \mathrm{null}(S)$. Is this true? :) Commented Aug 21, 2016 at 23:31
• I usually use $\subset$ for not necessarily strict inclusions and $\subsetneq$ when I want to indicate strict inclusion. That is, for me (and others) $A \subset B$ includes the possibility that $A=B$. You are right that $\mathrm{null}(T_0) = \mathrm{null}(T)$. In general, the null space of a restricted map will only be contained in the original null space, but yes here we have a equality. Commented Aug 22, 2016 at 0:15