Exercise 1224Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Surjective linear map and image

#surjectivity#image#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.

Hint

Find preimages of $(1,0)$ and $(0,1)$ .

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Preimages. $f(1,0,0)=(1,0)$ and $f(0,0,1)=(0,1)$ .

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

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

Rank-nullity check. $\dim\ker f+\dim\operatorname{Im} f=1+2=3$ . Correct.

Conclusion. $f$ is surjective with kernel of dimension $1$ .