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I have a question about a homework question so don't expect a full solution. Just if someone could tell me how I should approach this question. I'm not really sure what kind of manipulation is allowed.

"Let $V$ be a vector space. Show that for every three linear maps $A, B, C : V \rightarrow V$ we have $\operatorname{rk}(ABC)\leq \operatorname{rk}(B)$ "

So all I really know I can say for sure is that if I let $\dim(V)=n$ then $\operatorname{rk}(B)\leq n$ and same for any of the other maps. But no idea how I can compare them and what even $ABC$ is. Is it composition of maps? how can I compare them given such limited info about them?

J126
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user198177
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1 Answers1

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Hint

You can prove separately that $$\operatorname{rk}(BC)\le \operatorname{rk}(B)$$ and $$\operatorname{rk}(AB)\le \operatorname{rk}(B)$$ and then you combine the two results.

For the first inequality notice that since $C(V)\subset V$ then

$$BC(V)\subset B(V)$$ and then take the dimension. The second inequality is almost similar. Can you finish the proof?

  • So been reading up and working on it, thought might be easier to do rk(ABC)≤rk(BC) then rk(BC)≤rkB but fell like I haven't really learned enough about operations on abstract vector spaces to complete this proof. I cant take the dimension as I dont know anything more about A,B or C or V. All I can think is that the domain of ABC is the image of BC but how to get this in proof that follows I dont know. Dont even know how I notate it and make it indisputable that one is not larger than the other. – user198177 Jan 20 '15 at 23:35