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!
Best Answer
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 $Q(x_1,x_2) = y_1^Ty_2$.