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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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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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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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