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The following question is about a lower bound on the rank of a composition of functions given as a simple expression for the two terms of the sum involved in the inequality.

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

Is it true that $\def\rank{\operatorname{rank}}\rank(f_3 \circ f_2) + \rank(f_2 \circ f_1) \geq \rank(f_3 \circ f_2 \circ f_1) + \rank(f_2) $?

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2 Answers 2

up vote 2 down vote accepted

The Frobenius Inequality for matrices says $$R(AB)+R(BC)\le R(ABC)+R(B)$$ where $R$ is the rank (and we assume all products are defined), so maybe you have it backwards?

See also this recent m.se question on that topic.

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No, this is not true in general. The left-hand side may be $0$ while the right-hand side is not. For instance, take $V_i=\mathbb R^2$ and $f_i$ all equal and given by the matrix $\pmatrix{0&1\\0&0}$ in some basis. Then the composition of any two of the transformations is the zero map, with rank $0$, but $\rank f_2=1$.

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