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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?

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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 $E^*$.

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  • $\begingroup$ 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. $\endgroup$ – davyjones Aug 15 '14 at 13:51
  • $\begingroup$ @davyjones For sure there is, just the commutativity of two linear maps. But Im not good at abstract stuffs XD. By the way, if you like the answer, please approve it to give me a little credit. Thanks. $\endgroup$ – Troy Woo Aug 15 '14 at 14:02
  • $\begingroup$ Thanks all the same! I'll keep on thinking about it! $\endgroup$ – davyjones Aug 15 '14 at 14:23

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