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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)

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  • $\begingroup$ 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
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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.

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