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Lower bound involving the rank of the composition of linear transformations

The following question is about a lower bound on the rank of a composition of functions given that was orgionally stated incorectly in this post Lower bound involving the rank of the composition of linear transformations.

Consider finite-dimensional vector spaces $V_1,V_2, V_3,V_4$ and linear transformations of these spaces $f_1 : V_1 \rightarrow V_2$, $f_2: V_2 \rightarrow V_3$, $f_3: V_3 \rightarrow V_4$.

How do you prove $\def\rank{\operatorname{rank}}\rank(f_3 \circ f_2) + \rank(f_2 \circ f_1) \leq \rank(f_3 \circ f_2 \circ f_1) + \rank(f_2) $?

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    Didn't my answer to the other question tell you what to do about this one?2011-10-26
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    Thanks for the comment I was not aware of the Frobenius Inequality.2011-10-26
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    As Gerry had already provided you with the name of this inequality, a) it would have made sense to mention that name in this question; b) did you try to search for proofs of the inequality? I immediately found two proofs doing so ([here](http://en.wikipedia.org/wiki/Rank_(linear_algebra)#cite_note-1) and [here](http://books.google.com/books?id=jgEiuHlTCYcC&lpg=PA134&ots=8O0lMpbhP3&dq=frobenius%20inequality&pg=PA134#v=onepage&q=frobenius%20inequality&f=false)). If you don't understand something in those proofs it would make more sense to ask about that specifically than to start from scratch.2011-10-26
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    @Didier: It's not really a duplicate -- that question asks whether a false inequality is true; this one asks for a proof of the true inequality.2011-10-26
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    @joriki, right. As you properly explain above, this is more a case of duplicate (potential) *answer* than duplicate question. I mention that other similar cases on the site hint at a very odd (to me) phenomenon, which is that some OP simply do not read answers to their own posts (even some they accept).2011-10-26

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The restriction of $f_3$ followed by the projection induces a linear map $$ im(f_2)/im(f_2\circ f_1) \to im(f_3\circ f_2)/im(f_3\circ f_2\circ f_1)$$ (where $im$ is the image subspace) which is evidently surjective. The result follows.