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$A, B$ are linear map from $\mathbb{R}^{12} \to \mathbb{R}^{12}$ and dim$null(A) = 3$, dim$null(B) = 5$ what value could dim$null(AB)$ be?

I think it could be greater than or equal to 5 because $$ Ker(B) \subset \{x\in \mathbb{R}^{12} : Bx \in Ker A\} = Ker(AB) $$ but is there are better result we can get?

Jason
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$\ker(AB)$ can have any dimension between $5$ and $8$. As you have already observed, $\ker(B) \subset \ker(AB)$, so that $\dim \ker(AB) \geq 5$. On the other hand: using the rank nullity theorem, we have $$ \dim\operatorname{im}(AB) = \dim\operatorname{im}(A|_{\operatorname{im}(B)}) = \dim\operatorname{im}(B) - \dim \ker (A|_{\operatorname{im}(B)})\\ = \dim\operatorname{im}(B) - \dim [\ker (A) \cap \operatorname{im}(B)] \\\geq \dim\operatorname{im}(B) - \dim \ker(A) = 12 - \dim \ker(B) - \dim \ker (A) $$ So, we have $$ 12 - \dim \ker(AB) \geq 12 - (\dim \ker (B) + \dim \ker (A)) \implies\\ \dim \ker(AB) \leq \dim \ker(B) + \dim \ker(A) $$

Ben Grossmann
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  • If you're interested in examples for which all possible dimensions are attained, it suffices to consider diagonal matrices with $1$'s and $0$'s on the diagonal. – Ben Grossmann Apr 05 '17 at 06:22
  • can you explain what happened there? $\dim\operatorname{im}(A|{\operatorname{im}(B)}) = \dim\operatorname{im}(B) - \dim \ker (A|{\operatorname{im}(B)})\ = \dim\operatorname{im}(B) - \dim [\ker (A) \cap \operatorname{im}(B)] $? I haven't got nothing like $\operatorname{im}(A|_{\operatorname{im}(B)}) $ on my lecture but I want to do this proof https://math.stackexchange.com/questions/3066546/how-to-show-that-dim-kerab-le-dim-ker-a-dim-ker-b?noredirect=1&lq=1 –  Jan 08 '19 at 18:49
  • @VirtualUser $A|_{S}$ is the restriction of the linear map $A$ to the subspace $S$. – Ben Grossmann Jan 08 '19 at 18:54