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

Possible Duplicate:
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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## marked as duplicate by Did, Listing, Davide Giraudo, Mike Spivey, t.b.Nov 7 '11 at 5:14

Didn't my answer to the other question tell you what to do about this one? –  Gerry Myerson Oct 26 '11 at 6:08
Thanks for the comment I was not aware of the Frobenius Inequality. –  user7980 Oct 26 '11 at 6:18
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 and here). If you don't understand something in those proofs it would make more sense to ask about that specifically than to start from scratch. –  joriki Oct 26 '11 at 7:43
@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. –  joriki Oct 26 '11 at 7:46
@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). –  Did Oct 26 '11 at 8:21
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.