Proof of Binet-Cauchy identity through the polarization transformation This questions is motivated by exercise 3.7 in Steele's "The Cauchy-Schwarz Master Class." This is not a homework (I am trying to learn some math by myself) and I have already posted the question on math.stackexchange.com, where it did not receive any answer. 
https://math.stackexchange.com/questions/1484994/proof-of-binet-cauchy-identity-through-the-polarization-transformation. 
I hope my question is not too off-topic here and someone can help me.
Exercise 3.7 asks to prove the Cauchy-Binet identity:
$$(CB) \qquad <a,s><b,t>-<a,t><s,b>=\sum_{j<k}\begin{vmatrix}a_j & b_j\\ a_k & b_k\end{vmatrix}\begin{vmatrix}s_j & t_j\\ s_k & t_k\end{vmatrix},$$
where $<,>$ is the scalar product, starting from Lagrange identity:
$$(L) \qquad <a,a><b,b>-<a,b>^2=\sum_{j<k}\begin{vmatrix}a_j & b_j\\ a_k & b_k\end{vmatrix}^2.$$
The author suggests to use to this purpose the polarization transformation
$$T(f(u),v)=\frac{f(u+v)-f(u-v)}{4},$$
saying that it is useful to transform squares in products.
In the sketch of the solution then he says that if we fix $b$ and polarize $a$ with $s$, we obtain
$$<a,s><b,b>-<a,b><s,b>=\sum_{j<k}\begin{vmatrix}a_j & b_j\\ a_k & b_k\end{vmatrix}\begin{vmatrix}s_j & b_j\\ s_k & b_k\end{vmatrix},$$
and then if we fix $a$ and $s$ and polarize $b$ with $t$ he says that we obtain the Cauchy-Binet identity.
Here it is my attempt that does not arrive to the wished identity.
I start observing that $T(f(u),v)$ is linear in $f()$, i.e. $T(\alpha_1 f_1(u)+\alpha_2 f_2(u),v)=\alpha_1 T(f_1(u),v)+ \alpha_2 T(f_2(u),v)$.
Then, I observe that if $f(u)=g(u,u)$, where $g(x,y)$ is bilinear in $x$ and $y$, it holds:
$$T(f(u),v)=\frac{g(u,v)+g(u,v)}{2},$$
and if $g()$ is also symmetric ($g(x,y)=g(y,x)$) it holds
$$T(f(u),v)=g(u,v).$$
If $b$ is considered as a constant, Lagrange identity can be read as $h_1(a)+h_2(a)=h_3(a)$, where $h_i(a)=g_i(a,a)$ with $g_i(x,y)$ symmetric and bilinear. Applying $T(,s)$ to both sides we obtain indeed:
$$<a,s><b,b>-<a,b><s,b>=\sum_{j<k}\begin{vmatrix}a_j & b_j\\ a_k & b_k\end{vmatrix}\begin{vmatrix}s_j & b_j\\ s_k & b_k\end{vmatrix},$$
My difficulty is now with the second transformation, because only the first term on the LHS can be expressed as a symmetric and bilinear function ($\tilde g_1(x,y)=<a,s><x,y>$), but the other terms can be expressed as just  bilinear functions, leading then to a (correct) but different equality:
$$<a,s><b,t>-\frac{<a,b><s,t>+<a,t><s,b>}{2}=\sum_{j<k}\frac{1}{2}\left(\begin{vmatrix}a_j & b_j\\ a_k & b_k\end{vmatrix}\begin{vmatrix}s_j & t_j\\ s_k & t_k\end{vmatrix}+\begin{vmatrix}a_j & t_j\\ a_k & t_k\end{vmatrix}\begin{vmatrix}s_j & b_j\\ s_k & b_k\end{vmatrix}\right).$$
Any help to understand how Steele was expecting the reader  to use the polarization transformation is welcome.
 A: I shall not discuss whether the Cauchy-Binet formula is true or not. The question is whether it is implied by Lagrange identity. I think that the answer is negative.
Let me define the $4$-linear form $\phi(a,b,s,t)$ as the left-hand side of (CB), minus the right-hand side. By definition, we have
$$\phi(a,b,s,t)=-\phi(b,a,s,t)=-\phi(a,b,t,s)=\phi(s,t,a,b).$$
From (L), we know also
$$\phi(a,b,a,b)=0.$$
the Question amounts to proving that $\phi\equiv0$.
Using tensor product, there exists a unique bilinear form $\rho$ over $n\times n$ matrices, such that $\rho(a\otimes b,s\otimes t)=\phi(a,b,s,t)$. The asumptions are that
$$\rho(F,G)=-\rho(F^T,G)=-\rho(F,G^T)=\rho(G,F),\qquad \rho(a\otimes b,a\otimes b)=0.$$
The form $\rho$ is therefore symmetric, and its quadratic form $Q(F):=\rho(F,F)$ vanishes over the cone of rank-one matrices. It is a classical fact that $Q$ is a linear combination of $2\times2$-minors of $F$.
Now, take such a $2\times2$ minor $F_{ij}F_{kl}-F_{il}F_{kj}$. Bilinearize it : $b(F,G):=F_{ij}G_{kl}+F_{kl}G_{ij}-F_{il}G_{kj}-F_{kj}G_{il}$. Consider the form
$$\rho(F,G):=b(F,G)-b(F^T,G)-b(F,G^T)+b(F^T,G^T).$$
Such a form satisfies all the requirements, yet does not vanish identically when $F,G$ are rank-one matrices. This provides a counter-example to the strategy, suggested by Steele.
