0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There is a problem with your approach here : $\ker(f)$ and $\ker(g)$ are suspaces of different spaces ($V$ and $W$, respectively), so you can't take their sum. They have a direct sum, but it would not be a subspace of $V$ and thus it couldn't contain $\ker(g\circ f)$. $\endgroup$ – Arnaud D. Oct 30 '17 at 15:29
  • $\begingroup$ Oh ok! Thanks for pointing that out. I am now trying to prove the inequality using the basis of $V$. $\endgroup$ – model_checker Oct 30 '17 at 15:33
3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $g':\Im(f)\to W.$ $\endgroup$ – model_checker Oct 30 '17 at 16:00
  • $\begingroup$ The equality occurs when $\Im (f)\subset \ker(g).$ $\endgroup$ – model_checker Oct 30 '17 at 16:02
  • $\begingroup$ Is this correct? $\endgroup$ – model_checker Oct 30 '17 at 16:02
  • $\begingroup$ 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 $\ker(g)\subset\operatorname{im}(f)$. $\endgroup$ – Arnaud D. Oct 30 '17 at 16:04
  • $\begingroup$ 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 $X$. $\endgroup$ – model_checker Oct 30 '17 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.