What is the dimension of $T$?

Question

Let $U$, $V$ and $W$ be finite dimensional linear spaces (with dimensions $l$, $m$ and $n>0$, respectively) over field $\mathbb{F}$, and let $f\in{\rm Hom}$ $(U,V)$, $g\in{\rm Hom}$ $(U,W)$ such that ${\rm Ker} $ $f\subseteq{\rm Ker}$ $g$. Suppose that ${\rm dim}$ ${\rm Im}$ $f=t$. Define $$S=\{h\in{\rm Hom}\,(V,W)\mid g=h\circ f\}$$ and $$T=\{\alpha\in{\rm Hom}\,(V,W)\mid\exists h_1,h_2\in S, {\rm s.t.} \,\alpha=h_1-h_2\}.$$

Question: Prove that $T$ is a subspace of ${\rm Hom}(V,W)$ and determine the dimension of $T$.

I’ve managed to prove that $T$ is a subspace of ${\rm Hom}(V,W)$. I’m stuck when trying to figure out the dimension of $T$. Any help is appreciated.

Answer 1

The following observation is helpful.

Claim: $T = \{\alpha \in \operatorname{Hom}(V,W): \alpha \circ f = 0\}$

Proof: Let $T' = \{\alpha \in \operatorname{Hom}(V,W): \alpha \circ f = 0\}$; we wish to show that $T' = T$ (with $T$ defined as in the problem statement). It is easy to see that $T \subseteq T'$. Conversely, we can see that $T' \subseteq T$ as follows: let $\alpha$ be an element of $T'$. Let $h$ be any element of $S$ (noting that, due to the fact that $\ker f\subseteq \ker g$, $S$ is non-empty). Show that $h_1 = h+\alpha$ and $h_2 = h$ are two elements of $S$ for which $h_1 - h_2 = \alpha$. Thus, $\alpha \in T$. The conclusion follows. $\qquad\square$

Incidentally, this description allows you to efficiently show that $T$ is subspace.

Now, compute the dimension of $T$ with the help of this new description. It is helpful to note that $\alpha \in T$ if and only if $\operatorname{im}(f) \subseteq \ker(\alpha)$.