# possibly found a “counterexample” to a multilinear algebra problem

Edit: Exact Question. my question is b part $$\phi:E\times F\to G$$ 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

• 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 $H$. – MPW May 24 '19 at 17:53
• @mpw I'm not sure what $f\cdot g$ means, i don't have it anywhere – Anvit May 24 '19 at 18:04
• 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. – MPW May 24 '19 at 18:05
• @mpw I'm saying no linear function would exist $G\to H$ – Anvit May 24 '19 at 18:10
• @MPW Why do $G$ and $H$ need to be subsets of a field? $f\cdot g$ likely means composition of functions. – 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 $$N_1, N_2$$?