$$\newcommand{\im}{\mathrm{im}\;}\newcommand{\Span}{\mathrm{Span}}\newcommand{\rank}{\mathrm{rank}\;}$$For a bilinear map $\phi : V \times W \to E$ define the first nullspace as $$ N_1(\phi) = \{ v \in V : \forall w \in W . \phi(v,w) = 0 \}, $$ and the second nullspace as $$ N_2(\phi) = \{ w \in w : \forall v \in V . \phi(v,w) = 0\}. $$ It is obvious that resp. $N_1(\phi)$ and $N_2(\phi)$ are linear subspaces of resp. $V$ and $W$.

Let $\phi:V\times W \to E$, and $\psi:V\times W \to G$ be bilinear maps, such that $N_1(\psi) \subset N_1(\phi)$ and $N_2(\psi) \subset N_2$. W. H. Greub in his Multilinear Algebra book claims (1st ed., ch 1.1, problem 4b), that in this case there is a linear mapping $T : G \to E$, such that $\phi = T\circ\psi$.

I think, I found a counter example. Assume that $V,W,E,G$ are all standard Hilbert spaces over $\mathbb{R}$, with the inner product $\langle \cdot,\cdot \rangle$ and orthonormal base $(e_i)_{i=1}$, probably even of finite dimension. Let $$\psi(v,w) = \langle A(v),w \rangle e_1,$$ and $$ \phi(v,w) = B(v)f(w), $$ where $A : V \to W$ and $B : V \to G$ are linear with $\ker A \subset \ker B$ and $$ f : W \to \mathbb{R} $$ is a linear functional with $(\mathrm{im} \; A)^\bot \subset \ker f$. Thus, $ \ker A = N_1(\psi) \subset N_1(\phi) = \ker B$ and $ (\im A)^\bot = N_2(\psi) \subset N_2(\phi) = \ker f $. But in the case $\rank B > 1$, it is clear that $\dim \im \phi > \dim \mathrm{\im \psi} = 1 $ and it is impossible to linearly map a 1d subspace onto a multidimensional one.

To be concrete take $V,W,G,E = \mathbb{R}^4$, take $A$ to be an orthogonal projection to the subspace $\Span(e_1,e_2,e_3)$, and $B$ to be an orthogonal projection to $\Span(e_1,e_2)$, while $f(w) = \langle w, e_1 \rangle $.

This result contradicts the text, so I need some external justification. Am I right or am I wrong?


The statement, as you wrote it, is obviously false: in the particular case where $V=W$, $E=G=K$ (where $K$ is the base field, $\mathbb{R}$ if you want), and $N_1(\phi)=N_1(\psi)=N_2(\psi)=N_2(\psi)=0$, it would imply that two nondegenerated quadratic forms are always multiple of each other, which is nonsense. If you did not make a mistake when copying the statement, then the book is wrong.

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