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!

  • $\begingroup$ 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 $F^p$ $\endgroup$ Commented Jan 20, 2019 at 18:19
  • $\begingroup$ Ahh yes you are right. Dumn mistake on my part. I'm still very unsure how to prove it, do you have any ideas? $\endgroup$
    – TAPLON
    Commented Jan 20, 2019 at 18:21

1 Answer 1


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.


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