Understanding a bilinear form problem from Greub's Multilinear Algebra

I read the following problem from exercise sets of Greub's Multilinear Algebra, Chapter I, Sec. 1

Let $E$, $E^*$ be a pair of dual spaces and assume that $\mathit{\Phi}:E^{*}\times E\to\Gamma$ is a bilinear function such that $$\mathit{\Phi}(\tau^{*-1}x^{*},\tau x)=\mathit{\Phi}(x^{*},x)$$ for every pair of dual automorphisms. Prove that $\mathit{\Phi}(x^{*},x)=\lambda\langle x^{*},x\rangle$ where $\lambda$ is a scalar.

by my understanding, the pair of dual automorphisms refer to $\tau^{*}:E^*\to E^*$ and $\tau:E\to E$, hence the assumption on the bilinear function is equivalent to $$\mathit{\Phi}(\tau^{*}x^{*},x)=\mathit{\Phi}(x^{*},\tau x)\quad\forall\tau,\tau^*$$ which is very similar to the definition of $\tau^*$ (the dual mapping) from $\tau$ via non-degenerate bilinear form. Note the author use $\langle\cdot,\cdot\rangle$ to denote a non-degenerate bilinear function (in his linear algebra book).

Here is my question, I thought this exercise provides another characterization of non-degenerate bilinear forms up to a constant multiplication via exhausting every (dual) mappings, but it seems totally all right if I set $\mathit{\Phi}(x^*,x)=0$ $\forall x^*,x$.

How to understand this exercise correctly?

In coordinate space (after assigning basis and its dual basis to $E,E^*$), $\Phi$ can be thought of as a matrix $B$, $\tau$ an invertible matrix $A$, and $\tau^*$ the matrix $A^T$. The given condition is equivalent to: $$A^{-1}BA=B, \forall A\quad\text{ or }\quad AB=BA, \forall A$$ Obviously $B$ has to be the scalar multiple of identity matrix, which is the dual pairing $\langle,\rangle$ under a set of basis for $E$ and its dual basis for $E^*$.
• Wow! It is so enlightening this way! I understand the purpose of the exercise now. BTW, is there a coordinate-free way to express this $AB=BA$ thing? Much appreciated. – davyjones Aug 15 '14 at 13:51