# Equality of Frobenius Inequality

Consider the case when $$\DeclareMathOperator{\Rank}{Rank}\Rank(AB) + \Rank(BC) = \Rank(B) + \Rank(ABC)$$

where $$A \in M_{m,k}(F)$$ , $$B \in M_{k,p}(F)$$ and $$C \in M_{p,n}(F)$$ for any field $$F$$.

I wish to show that $$\DeclareMathOperator{\Ker}{Ker}\Ker(AB) \subseteq \Ker(B) + \operatorname{Range}(C)$$

I have been stuck on this problem for a while now. In general the problem statement doesn't even make sense to me, it seems to me that $$\Ker(AB)$$ is a subset of $$F^m$$, $$\Ker(B)$$ is a subset of $$F^k$$ and $$\operatorname{Range}(C)$$ is a subset of $$F^n$$. So I don't see how we could have such a containment if the dimensions don't line up.

Any help is appreciated thanks!

• It looks like you might have mixed up the domain and codomain of your maps. Usually $M_m,k(F)$ is the set of linear maps from $F^m$ to $F^k$. The kernel of a map is a subset of the domain. Both $B$ and $AB$ act on $F^p$ so their kernels are part of $F^p$. Similarly, the image of $C$ is in $F^p$ Jan 20, 2019 at 18:19
• Ahh yes you are right. Dumn mistake on my part. I'm still very unsure how to prove it, do you have any ideas? Jan 20, 2019 at 18:21

If $$T:V\to W$$ is a linear mapping, then dimension theorem says $$\dim V =\dim{N}(T)+\dim R(T)$$ where $${N}(T)$$ is the null space of $$T$$ and $$R(T)$$ is the range of $$T$$.
Given linear maps $$T:V\to W$$ and $$S:W\to X$$, we can apply dimension theorem to $$S|_{R(T)}:R(T)\to X$$ and get $$\dim R(T) = \dim N(S|_{R(T)}) +\dim R(S|_{R(T)})=\dim \left[N(S)\cap R(T)\right] +\dim R(ST)$$or equivalently $$\dim R(T)-\dim R(ST)=\dim \left[N(S)\cap R(T)\right].$$
The given equation can be written as $$\DeclareMathOperator{\Rank}{Rank}\Rank(B)-\Rank(AB) = \Rank(BC) -\Rank(ABC)$$ or $$\dim R(B)-\dim R(AB) =\dim R(BC)-\dim R(ABC).$$ As we've already seen, it is equivalent to $$\dim [N(A)\cap R(B)]=\dim [N(A)\cap R(BC)].$$ Since $$N(A)\cap R(B)\ge N(A)\cap R(BC)$$, it says that the two spaces are equal.
Assume $$v\in N(AB)$$. Then $$ABv=0$$ implies that $$Bv\in N(A)\cap R(B)=N(A)\cap R(BC).$$ Thus there exists $$w$$ such that $$Bv = BCw$$. Now, since $$B(v-Cw)=0$$, there exists $$x\in N(B)$$ such that $$v-Cw =x$$. This gives $$v= x +Cw \in N(B)+R(C),$$ proving $$N(AB)\le N(B)+R(C)$$ as desired.