0
$\begingroup$

Let $f: U\longrightarrow V$ and $g: V \longrightarrow W$ be linear maps. If $g\circ f$ is surjective, then ker(g)+im(f)=V. Is there an easy way to see/proof this? I managed to show it by first showing

dim(ker g $\cap$ im f) = dim ker(g $\circ$ f) − dim ker f

and then using the Rank-Nullity Theorem and dimension arguments, but that seems to be to much work.

$\endgroup$
2
  • $\begingroup$ Just out of curiosity: Why are you not happy with this solution? $\endgroup$ Sep 30 '20 at 8:03
  • $\begingroup$ Because for showing the first result one uses 2 times the Rank-Nullity Theorem and after that again one time. So it seems to be just far from the shortest/direct way to prove it. $\endgroup$
    – RobbiTobbi
    Sep 30 '20 at 8:07
1
$\begingroup$

The dimension argument is ok as long as you assume that $V$ is finite dimensional. Otherwise it is wrong (you cannot subtract cardinal numbers, and there's no way around it).

But the statement is true for any $V$.

Proof. Let $v\in V$. Since $g\circ f$ is surjective, then

$$(g\circ f)(u)=g(v)\text{ for some }u\in U$$

therefore $g(f(u)-v)=0$. In particular $f(u)-v\in\ker(g)$. Meaning

$$v=f(u)+v_g\text{ for some }v_g\in\ker(g)$$

By the arbitrary choice of $v$ we conclude that $V=\text{im}(f)+\ker(g)$. $\Box$

$\endgroup$
2
  • $\begingroup$ Thank you, V finite dimensional was fine for my application, but this is much more straight forward. $\endgroup$
    – RobbiTobbi
    Sep 30 '20 at 8:31
  • 1
    $\begingroup$ Just a side note: Instead of introducing another variable, we could just write $v=f(u)+(f(u)-v)$ and note that $f(u)\in\operatorname{im}(f)$ and $f(u)-v\in\operatorname{ker}(g)$. This gets rid of the unspecified "some $v_g$", since we pinned down what this $v_g$ is anyway. $\endgroup$
    – Christoph
    Sep 30 '20 at 8:35

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.