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"
$F=\mathbb R^3$
$H=E=\mathbb R^2$
$G=\mathbb R^5$

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

  • $\begingroup$ 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

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 $N_1, N_2$?

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  • $\begingroup$ 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

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