Exercise 128Linear algebra › Linear mapsL1Added 3 Nov 2026

An injective linear map

#injectivity#kernel#linear map

Statement

Let $f:\mathbb{R}^2\to\mathbb{R}^3$ be defined by $f(x,y)=(x+y,\ x-y,\ 2x)$ . Show that $f$ is injective.

Hint

A linear map is injective if and only if its kernel is $\{0\}$ (because $f(u)=f(v)\iff f(u-v)=0$ ). Compute $\ker f$ .

Show solutionHide solution

Solution

Kernel. Solve $x+y=0$ , $x-y=0$ , $2x=0$ . Adding the first two equations gives $2x=0$ , so $x=0$ , and then $y=0$ .

Conclusion. $\ker f=\{0\}$ , so $f$ is injective. By rank-nullity, $\operatorname{rank}(f)=2-0=2$ , which agrees with the plane found as the image of this same map in the exercise on computing an image.

An injective linear map

Statement

Solution

More problems in Linear maps