Exercise 130Linear algebra › Linear mapsL1Added 3 Nov 2026

A surjective linear map

#surjectivity#rank#linear map

Statement

Let $f:\mathbb{R}^3\to\mathbb{R}^2$ be defined by $f(x,y,z)=(x+y,\ y+z)$ . Show that $f$ is surjective, and compute $\ker f$ .

Hint

Find a preimage of an arbitrary vector $(a,b)$ .

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Surjectivity. For any $(a,b)\in\mathbb{R}^2$ , $f(a,0,b)=(a+0,\ 0+b)=(a,b)$ . So $f$ is surjective.

Kernel. $x+y=0$ and $y+z=0$ give $y=-x$ and $z=-y=x$ , so $\ker f=\operatorname{span}\big((1,-1,1)\big)$ .

Conclusion. $f$ is surjective, $\operatorname{rank}(f)=2$ , $\dim\ker f=1$ , and $2+1=3=\dim\mathbb{R}^3$ .