# All nondegenerate bilinear symmetric forms on a complex vector space are isomorphic

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!

• What do you mean by "isomorphic"? Aug 16, 2020 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.
– user637978
Aug 16, 2020 at 12:42

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

• Is the same true for a real vector space?
– user637978
Mar 26, 2021 at 15:58
• @MutatedPenguin No. For a real vector space, Sylvester's law of inertia applies. Mar 26, 2021 at 17:48