3
$\begingroup$

Edit: Exact Question. my question is b part enter image description here$\phi:E\times F\to G$ be bilinear
$\psi:E\times F\to H$ be bilinear
Given $N_1(\phi)\subset N_1(\psi)$ and $N_2(\phi)\subset N_2(\psi)$ Show that there exist a linear function $f:G\to H$ such that $f\cdot\phi=\psi$ where $N_1(\phi) = \{x|\phi(x,y)=0\forall y\in F\}$ and similarly $N_2$ for the second coordinate

My "counterexample"
$F=\mathbb R^3$
$H=E=\mathbb R^2$
$G=\mathbb R^5$
$\phi(e,f)=(e_1f_1,e_1f_2,e_1f_3,e_2f_1,e_2f_2)$
$\psi(e,f)=(e_2f_2,e_2f_3)$
$N_1(\phi)=\{(0,0)\}\subset\{(x,0)\}=N_1(\psi)$
$N_2(\phi)=\{(0,0,0)\}\subset\{(x,0,0)\}=N_2(\psi)$

Please point out the mistake in the "counterexample". Also please provide hints for the problem

$\endgroup$
  • $\begingroup$ It seems as if $G$ and $H$ should be subsets of $\mathbb R$ (or some other field). Otherwise what does $f\cdot g$ mean? $f$ and $\psi$ both take value in $H$. $\endgroup$ – MPW May 24 '19 at 17:53
  • $\begingroup$ @mpw I'm not sure what $f\cdot g$ means, i don't have it anywhere $\endgroup$ – Anvit May 24 '19 at 18:04
  • $\begingroup$ Huh? Then how can you have a counterexample? A counterexample would be a scenario satisfying the hypotheses but for which no such linear function $f:G\to H$ exists. $\endgroup$ – MPW May 24 '19 at 18:05
  • $\begingroup$ @mpw I'm saying no linear function would exist $G\to H$ $\endgroup$ – Anvit May 24 '19 at 18:10
  • $\begingroup$ @MPW Why do $G$ and $H$ need to be subsets of a field? $f\cdot g$ likely means composition of functions. $\endgroup$ – Michael Burr May 24 '19 at 18:13
1
$\begingroup$

Your counterexample is correct, it shows the claim made before cannot be correct (there is no way to get $e_2f_3$ as a linear combination of the components of $\phi(e,f)$). Are you sure you copied the problem correctly, especially the definition of $N_1, N_2$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have added the exact question as well. I think I've copied correctly $\endgroup$ – Anvit May 24 '19 at 20:03
  • $\begingroup$ Yes, I think so as well. I'm at a loss and maybe somebody else can shed some light onto this. $\endgroup$ – Ingix May 24 '19 at 21:30

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.