# How do I prove that $\operatorname{rank}{AB}+\operatorname{rank}{BC}\le\operatorname{rank}{B}+\operatorname{rank}{ABC}$? [duplicate]

The Frobenius inequality of linear algebra, with $A,B,C\in M_n(\mathbb{F})$, is: $$\operatorname{rank}{AB}+\operatorname{rank}{BC}\le\operatorname{rank}{B}+\operatorname{rank}{ABC}$$

-

## marked as duplicate by Gerry Myerson, rschwieb, William, Norbert, J. M.Oct 7 '12 at 13:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

In wikipedia, I've already read the proof you referred to. What I don't understand is, what is $\ker{ABC}/\ker{BC}$ and why that projection is one-to-one? – Voldemort Sep 19 '12 at 0:03

## 1 Answer

To answer your more focused question in the comment:

The clear candidate map is $x\mapsto Cx +\ker (B)$, which gives you a homomorphism from $\ker(ABC)\rightarrow \ker(AB)/\ker(B)$.

Modding out by this map's kernel, which happens to be $\ker(BC)$, the new map from $\ker(ABC)/\ker(BC)\rightarrow \ker(AB)/\ker(B)$ is necessarily an injection. (This is just an isomorphism theorem.)

I'll also add that there is no indication that it is (or even needs to be) onto, so it's not really a projection.

From that point on, presumably you are working with all finite dimensional spaces, so this isomorphism gives you the inequality $\dim(\ker(ABC))-\dim(\ker(BC))\leq \dim(\ker(AB))-\dim(\ker(B))$

-