TITLE: $f,g:\mathbb R^3\to\mathbb R$ are nonzero linear maps. Then which are true?
QUESTION [5 upvotes]: (A) is wrong: Consider $f:(x,y,z)\mapsto(x,0,0),~g:(x,y,z)\mapsto(x,y,0)$
(B) is wrong: Consider $f:(x,y,z)\mapsto(x,0,0),~g:(x,y,z)\mapsto(x,y,0)$
(C) is wrong: $\ker g$ is a subspace of $\mathbb R^3$ and let $f:\ker g\to\mathbb R$ be the restriction of $f.$ Then since $\ker f\subset\ker g$ by isomorphism theorem $\ker g /\ker f\simeq\Im(f)\ne\mathbb R.$
(D) is correct.

Please tell me if I'm right. I'm skeptical especially about (C).

REPLY [6 votes]: $f,g:\mathbb{R}^3\to \mathbb{R}$ are required to be non-zero linear maps into $\mathbb{R}$ so your counterexamples for (A) and (B) (in particular, the map $g:\mathbb{R}^3\to \mathbb{R}^2$) are not valid. 
The key point is to use the rank-nullity theorem to note that $\text{ker}(f)$ and $\text{ker}(g)$ are two-dimensional subspaces of $\mathbb{R}^3$ with one contained in the other; what does this imply? 
I hope this helps!