I want to show that

$$ \dim \ker(AB) \le \dim \ker A + \dim \ker B. $$

My problem

I thought that I can do that in this way:
Let consider $x \in\ker B$ $$Bx = 0$$ Let multiplicate this from left side by A and we get: $$ABx = 0$$ so $$ker B \subset\ker AB $$ so $$\dim \ker(B) \le \dim\ker AB$$

We can do the same thing with $\ker A$

let consider $ \vec{y} \in \operatorname{im}(AB) $ so $$ y = (AB)x $$ what is equivalent to $$ \vec{y} = A(B\vec{x}) = A\vec{w} $$ So $$ \vec{y} \in \operatorname{im}(AB) \rightarrow \vec{y} \in \operatorname{im}(A)$$ so $$ \operatorname{rank} AB \le \operatorname{rank} A \leftrightarrow \dim \ker A \le \dim \ker AB $$ But I am not sure what I should do later...


I have seen this post $A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ but I haven't got nothing like $\operatorname{im}(A|_{\operatorname{im}(B)})$ on my algebra lecture and I can't use that so I search for another proof (or similar without this trick)

  • $\begingroup$ Perhaps see here? math.stackexchange.com/questions/269474/… $\endgroup$ – T. Fo Jan 8 at 18:59
  • $\begingroup$ @T.Ford I have suggested Sugata Adhya's post but on finish he failed, Mikko Korhonen used what I can't and Babak Miraftab doesn't response to comment, which represents my doubts too :( So I thought that it can be proved in similiar way as I presented in post $\endgroup$ – VirtualUser Jan 8 at 19:01
  • $\begingroup$ You have this backwards. Since $\ker B\subset \ker(AB)$, we have $\dim\ker B \le \dim\ker(AB)$, not the other way around. $\endgroup$ – Ted Shifrin Jan 8 at 20:20

This is a proof in general where $A:V\to W$ and $B:U\to V$ are linear maps. Here $U$, $V$, and $W$ are arbitrary vector spaces over a base field $F$, and they do not necessarily have finite dimensions. That is, $$\dim \ker (AB) \leq \dim \ker A+\dim\ker B$$ is true whether or not the relevant dimensions are finite cardinals.

Note that $x\in \ker(AB)$ iff $Bx\in \ker A$, which is the same as saying $$x\in B^{-1}(\ker A\cap \operatorname{im}B).$$ Recall from the isomorphism theorems that $\operatorname{im} B\cong U/\ker B$ so there exists an isomorphism $$\varphi: U\overset{\cong}{\longrightarrow} \ker B\oplus \operatorname{im}B.$$ In other words, $$\varphi\big(B^{-1}(\ker A\cap \operatorname{im}B)\big)=\ker B\oplus (\ker A\cap \operatorname{im}B).$$ Consequently, \begin{align}\dim\ker(AB)&=\dim\big(\ker B\oplus (\ker A\cap \operatorname{im}B)\big)\\&=\dim\ker B+\dim(\ker A\cap \operatorname{im}B).\end{align} Since $\ker A\cap \operatorname{im}B\subseteq \ker A$, we obtain the desired inequality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.