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All nondegenerate bilinear symmetric forms on a complex vector space are isomorphic. Does this mean that given a nondegenerate bilinear symmetric forms on a complex vector space that you can choose a basis for the vector space such that the matrix representation of the bilinear form is the identity matrix? Can somebody help explain to me why this is?

I'm thinking that a matrix with entries in $\mathbb{C}$ is going to have a characteristic equation that splits into linear factors (with multiplicities) and so will be diagonalizable, but still can't quite put these pieces together. Insights appreciated!

Vincenzo Tibullo
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    What do you mean by "isomorphic"? – Marktmeister Aug 16 '20 at 12:16
  • Hmm.. Good question. I think I mean that their matrix representations can be made identical by choosing a proper basis, and perhaps this implies that the matrix representations can both be made to be the identity. –  Aug 16 '20 at 12:42

1 Answers1

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The answer is yes.

First, a proof that the bilinear forms are isomorphic. Note that it suffices to prove that this holds over $\Bbb C^n$.

First, I claim that every invertible, complex, symmetric matrix can be written in the form $A = M^TM$ for some complex matrix $M$. This can be seen, for instance, as a consequence of the Takagi factorization.

Now, let $Q$ denote a symmetric bilinear form over $\Bbb C^n$, and let $A$ denote its matrix in the sense that $Q(x_1,x_2) = x_1^TAx_2$. Let $Q_0$ denote the canonical bilinear form defined by $Q_0(x_1,x_2) = x_1^Tx_2$. We write $A = M^TM$ for some invertible complex matrix $M$.

Define $\phi:(\Bbb C^n, Q) \to (\Bbb C^n, Q_0)$ by $\phi(x) = Mx$. It is easy to verify that $\phi$ is an isomormphism of bilinear product spaces, so that the two spaces are indeed isomorphic.

With all that established: we can see that the change of basis $y = Mx$ is such that $Q(x_1,x_2) = y_1^Ty_2$.

Ben Grossmann
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