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Suppose $f:X\times Y\rightarrow Z$ and $g:X\times Y\rightarrow W$ are bilinear maps in the category of vector spaces (say, real). Define the null space $N_1(f) :=\{x \in X: f(x,y)=0 \ \forall\, y\in Y\}$ and $N_2(f)$ analogously. If it's known that $N_i(f)\subset N_i(g)$, $i=1,2$, then how can show the existence of a linear map $L:Z\rightarrow W$ such that $L\circ f = g$?

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migrated from Jul 23 '13 at 12:06

This question came from our site for professional mathematicians.

In general, there is no such $L$. Consider, as a simple example, $X = Y = Z = W = \mathbb{C}$, viewed as a vector space over $\mathbb{R}$. Let

$$f(x,y) = x\cdot y;\quad g(x,y) = x\cdot \overline{y}.$$

Evidently, $N_i(f)$ and $N_i(g)$ are all trivial, but $L(x) = L(f(x,1)) = g(x,1) = x$ forces $L(z) = z$, and $L(y) = L(f(1,y)) = g(1,y) = \overline{y}$ requires $L(z) = \overline{z}$.

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