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The following problem is an unsolved practice problem from a textbook which is assigned at the end of the chapter. So, I need hints on how to start solving the problem below:

Show that if $U$, $V$, $W$ are finite dimensional vector spaces, and $f\in \mathrm{Hom}\left ( U,V \right )$, $g\in \mathrm{Hom}\left ( V,W \right )$, then: $\dim \mathrm{Ker}\left ( gf \right )\leqslant \dim \mathrm{Ker}(f) + \dim \mathrm{Ker}(g)$

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    Please (i) provide context (is this homework? Self-study? In what context did you find the problem?) (ii) say a few words about what you've tried or what your thoughts are; and (iii) try not to simply write a problem as if you were assigning homework to the group; some of us find that at least mildly rude. The first two points help ensure that you receive answers that are at the appropriate level, and that people don't waste time suggesting things you've already thought about.2012-01-20
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    I think you want to write $g \circ f$.2012-01-20
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    Are your functions written on the right? Otherwise, $fg$ does not make sense.2012-01-20
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    @Arturo Magidin: You're right. The functions are written $gf$ and not $fg$. I have already fixed the original statement.2012-01-20
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    @m_p2009: That's not just information for me: that's information that is relevant to **all** readers. Please put it in the post, not the comments, and in the future please try to write in the context into the post from the very first.2012-01-20

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Note: my functions are written on the left of their argument.

Hint the first: $\mathrm{ker}(f)\subseteq \mathrm{ker}(g\circ f)$.

Hint the second: Let $U' = \{ u\in U\mid f(u)\in\mathrm{ker}(g)\}$. Now use the Rank-Nullity Theorem.

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    I can prove the first hint. For the second one, In general if $T$ is a linear map from $V$ to $W$, then the rank nullity theorem is given by: $dim Ker T + dim Im T = Dim V$. In your case, how are you applying the rank nullity theorem? and to which spaces?2012-01-20
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    @m_p2009: To the restriction of $f$ to $U'$, which is a linear map from $U'$ to $\mathrm{ker}(g)$; and note that the Rank-Nullity Theorem tells you that $$\dim V = \dim\mathrm{ker}(T) + \dim\mathrm{Im}(T) \leq \dim\mathrm{ker}(T)+\dim(W).$$2012-01-20
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    Note that $U'$ is **precisely** the kernel of $gf$. The image is a *subspace* of the kernel of $g$. The kernel is the kernel of $f$ (by the first hint).2012-01-20
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    This is what I can deduce based on the previous hints $Dim \left ( U^{'}\right )=dim\left ( Ker gf \right )\leqslant dim Ker (f/U^{'})+dim Ker\left ( g \right )$. Can we say that $dim Ker (f/U^{'}) \leqslant dim Ker (f)$?2012-01-20
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    @m_p2009: as the last sentence of my last comment notes, $\mathrm{ker}(f|_{U'}) = \mathrm{ker}(f)\cap U' = \mathrm{ker}(f)$. So you have *equality* of subspaces, hence of dimensions.2012-01-20