Exercise 140Linear algebra › Linear mapsL1Added 3 Nov 2026

Kernel of a composition

#composition#kernel#inclusion

Statement

Let $f:E\to F$ and $g:F\to G$ be linear maps. Show that $\ker f\subseteq\ker(g\circ f)$ , with equality if and only if $\ker g\cap\operatorname{Im} f=\{0\}$ (that is, if and only if the restriction of $g$ to $\operatorname{Im} f$ is injective). In particular, equality holds if $g$ is injective.

Hint

Note that $\ker(g\circ f)=\{x\in E: f(x)\in\ker g\}$ .

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Solution

Inclusion. If $x\in\ker f$ , then $f(x)=0$ , so $(g\circ f)(x)=g(0)=0$ .

Equality if $\ker g\cap\operatorname{Im} f=\{0\}$ . Let $x\in\ker(g\circ f)$ . Then $f(x)\in\ker g$ and $f(x)\in\operatorname{Im} f$ , so $f(x)\in\ker g\cap\operatorname{Im} f=\{0\}$ , i.e. $x\in\ker f$ .

Converse. Suppose $\ker f=\ker(g\circ f)$ and let $w\in\ker g\cap\operatorname{Im} f$ , say $w=f(x)$ . Then $g(f(x))=g(w)=0$ , so $x\in\ker(g\circ f)=\ker f$ , hence $w=f(x)=0$ .

Special case. If $g$ is injective, $\ker g=\{0\}$ , so $\ker g\cap\operatorname{Im} f=\{0\}$ .

Conclusion. $\ker f\subseteq\ker(g\circ f)$ , with equality exactly when $\ker g\cap\operatorname{Im} f=\{0\}$ .

Kernel of a composition

Statement

Solution

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