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Prove that $\operatorname{rank} (f) + \operatorname{rank} (g) -\dim W\leq \operatorname{rank}(g\circ f)$, where $f:V\to W$ and $g:W\to X$ with both $f$ and $g$ being linear maps.

My attempt: The inequality is equivalent to proving that $\dim \ker (f)+\dim \ker(g)\geq \dim \ker(g\circ f).$ To prove this we can show that the set $\ker(g\circ f )\subset \ker(f)+\ker(g).$ Let $\ker(f)+\ker(g)=\{v+w:f(v)=0\text{ and } g(w)=0\}.$ So let $x\in V$ be an element of the $\ker(g\circ f).$ Then $g(f(x))=0.$ If $x\in \ker f$ then $f(x)=0_W$ and so we can represent $x$ as $x+0_W\in \ker(f)+\ker(g).$ I am not sure how to proceed further.

I know why this inequality must be true. There are some elements in $W$ not equal to $0$ that are included in the kernel of $g.$ If even one does not have a pre-image in $V$ then we get a strict inequality and if all them have a preimage then we get an inequality. I am just not able to verbalize this in a formal proof. Any insights would be much appreciated.

Michael Hardy
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Student
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1 Answers1

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Hint : The restriction of $g$ to the image of $f$ is a linear map $\operatorname{im}(f)\to W$, whose image is the image of $g\circ f$, and whose kernel is $\ker(g)\cap \operatorname{im}(f)$. What does the rank-nullity theorem tells you for this linear map?

Arnaud D.
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  • So we have $\text{rank} (f)=\dim \ker (g')+\dim \Im (g')\leq \dim \ker (g)+\text{rank} (g\circ f )\leq \dim W-\text{rank}(g)+\text{rank}(g\circ f).$ And so we are done. Here $g':\Im(f)\to W.$ – Student Oct 30 '17 at 16:00
  • The equality occurs when $\Im (f)\subset \ker(g).$ – Student Oct 30 '17 at 16:02
  • Is this correct? – Student Oct 30 '17 at 16:02
  • You got it (although the second inequality in your first comment is actually an equality). For the equality, it should be the reverse inclusion; the equality occurs when $\ker(g)\cap\operatorname{im}(f)=\ker(g)$, which is equivalent to $\ker(g)\subset\operatorname{im}(f)$. – Arnaud D. Oct 30 '17 at 16:04
  • But why is $\dim \ker(g') =\dim \ker (g)$? It could be the case that I have elements in $W$ not included in the image of $f$ that are mapped to the zero element by $g$ in $X$. – Student Oct 30 '17 at 16:13
  • Of course it could happen. But then, you have a strict inclusion $\ker(g')\subset \ker(g)$, thus a strict inequality between their dimension. – Arnaud D. Oct 30 '17 at 16:16
  • I think that $\dim \ker (g')\leq \dim \ker (g).$ When the equality occurs we have that $\dim \ker (g')=\dim \ker (g)$ and thus $\ker (g)\cap \Im(f)=\ker (g).$ So we should have $\ker (g)\subset \Im (g)$. (I made an error and I see why my previous comment was wrong). Thanks for pointing it out! – Student Oct 30 '17 at 16:17
  • That is correct. – Arnaud D. Oct 30 '17 at 16:18