Exercise 1217Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Injective linear map and kernel

#injectivity#kernel#linear map

Statement

Let $f:E\to F$ be a linear map. Show that $f$ is injective if and only if $\ker f=\{0\}$ .

Hint

For the forward direction, use $f(0)=0$ . For the converse, assume $f(u)=f(v)$ and use linearity.

Show solutionHide solution

If $f$ is injective. $f(0)=0$ because $f$ is linear. If $u\in\ker f$ , then $f(u)=0=f(0)$ , so $u=0$ by injectivity. Hence $\ker f=\{0\}$ .

Conversely. Suppose $\ker f=\{0\}$ . If $f(u)=f(v)$ , then $f(u-v)=0$ , so $u-v\in\ker f=\{0\}$ , i.e. $u=v$ . So $f$ is injective.

Conclusion. $f$ is injective iff $\ker f=\{0\}$ .