# How to show that $\dim\ker(AB) \le \dim \ker A + \dim \ker B$?

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...

### edited

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)

• Perhaps see here? math.stackexchange.com/questions/269474/… – T. Fo Jan 8 at 18:59
• @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 – VirtualUser Jan 8 at 19:01
• You have this backwards. Since $\ker B\subset \ker(AB)$, we have $\dim\ker B \le \dim\ker(AB)$, not the other way around. – 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.