# Composition of linear mappings is $0$ if and only if $\operatorname{im}(g) \subseteq \ker(h)$

Let $$U,V,W$$ be vector spaces and $$g:U\rightarrow V,h:V \rightarrow W$$ be linear mappings. Prove that the composite linear mapping $$h\circ g:U\rightarrow W$$ satisfies $$h\circ g=0$$ if and only if $$\operatorname{im}(g) \subseteq \ker(h)$$.

This statement seems obvious but I am terrible at proving things so I have no idea how to even begin.

• Hint: What dies it mean that $\;(h\circ g)(x)=0$? – Bernard May 5 at 10:05

## 1 Answer

I will show you one direction: Let $$\text{im}(g) \subseteq \text{ker}(h)$$. For $$u \in U$$, we have $$g(u) \in \text{im}(g)$$, so $$h(g(u)) = 0$$ since $$\text{im}(g) \subseteq \text{ker}(h)$$. Try the other direction, you can do it.

• Let $h\circ g=0$. This means that for all $v \in \operatorname{im}(g), h(v)=0$ which implies that $\operatorname{im}(g) \subseteq \ker(h)$? Is this correct? – Montes May 5 at 10:14
• yap, good job!! – Riquelme May 5 at 10:49