$f(x)=0$ implies $f(g(y))=0$ for some $y$?

If $f,g : V \to V$ are linear maps on a (edit: finite dimensional) vector space $V$ satisfying $f(x)=0$ for some nonzero $x$, does it follow that $f(g(y))=0$ for some nonzero $y$? Wh$y$?

This is part of a larger proof that I'm stuck on. I'd prefer hints over full solutions, since this is part of a homework problem.

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 It depends on what $g$ is. Take $f:\mathbb{R}^2\rightarrow\mathbb{R}$ with $f(x,y)=x$ and $g:\mathbb{R}\rightarrow\mathbb{R}^2$ with $g(x)=(x,0)$. – wj32 Oct 10 '12 at 3:53 Oops! I missed a constraint. I'll add it to the question. – Pink Oct 10 '12 at 3:54 The only way this would be true all of the time is if $x=g(y)=0$ since every linear map contains the $0$ vector. – Jeremy Oct 10 '12 at 3:57 @Pink: We need to ensure $g(y) \ne x$ for all $y \in V$. What does this imply about the image of $g$? Supposing that $V$ is finite-dimensional, what does this imply about the kernel of $g$? – wj32 Oct 10 '12 at 4:00 @wj32: Suppose g(0)=0=f(0) By the Wikipedia article this is required. – Ross Millikan Oct 10 '12 at 4:07

Let $h=f\circ g$; if $\ker h=\{0_V\}$, then $h$ is invertible. But $\ker f\ne\{0_V\}$, so $f$ is not invertible. Therefore?

Added: I should have said that I was assuming that $V$ was finite-dimensional; fortunately, it was.

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 Therefore $f \circ g = h$ is not invertible, meaning $\ker h \supset \{0_V\}$ (strict '$\supset$'). Thanks! – Pink Oct 10 '12 at 4:18 @Pink: Youv’e got it. You’re welcome. – Brian M. Scott Oct 10 '12 at 4:19 Perhaps this is a better question for meta, but should I be citing your hint on my homework? (I guess this is dependent on the professor. Suppose you were my professor--would you expect a student to cite a Stack Exchange hint?) – Pink Oct 10 '12 at 4:32 @Pink: As you say, it depends on the professor. It also depends on the hint. This is the sort of hint that I might well have given one of my own students, so it wouldn’t bother me. If you’ve no clear guidelines, I’d say that the safest thing this time would be to cite it word for word and ask your instructor what he/she thinks about such things. – Brian M. Scott Oct 10 '12 at 4:38 You need finite dimensions (or other condition) to ensure that invertibility of $h$ implies invertibility of $f$ (well, the other way around, but it is very awkward to state). – copper.hat Oct 10 '12 at 5:12
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Hint: If a linear map has to take $0$ to $0$... If not, the points that $f$ takes to $0$ could be a line the misses the $y$ such that $g(y)=0$.

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Yes, this does follow that $g(y) = 0$ for some non-zero $y$, and here is why.

If $g(y) = x$ for some $y$, then we're done because we know $f(x) = 0$.

If not, (assuming $V$ is finite dimensional) then $g$ has a non-trivial kernel so again we're done.

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(I'm assuming that $\dim V < \infty$.)

Consider the set $\Sigma = \{ z | Gz = x \}$. If $\Sigma$ is not empty, then you are finished. If $\Sigma$ is empty, what can you say about $\ker G$?

To illustrate why I have restricted the dimension of $V$, consider the infinite dimensional space $V = \{ (x_1,...) | x_i \in \mathbb{R} \}$. Then let $g((x_1,...)) = (0,x_1,...)$, and $f((x_1,...)) = (x_2,...)$ (ie, a right shift and left shift respectively). Then we have $f((1,0,...)) = 0$, but $f \circ g (x) = x$ for all $x \in V$.

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