# signature of biliner form and the signature on the induced Hermitian form

Let $$V$$ be finite dimension vector space over $$\Bbb{R}$$, and assume that $$b:V\times V \to \Bbb{R}$$ is the symmetric bilinear form, (therefore all eiginvalue are reals)

Then we can define the associated Hermitian form to be $$h:V_\Bbb{C} \times V_\Bbb{C} \to \Bbb{C}$$ such that $$h(x,y) = b(x,\bar{y})$$ (which is Hermitian symmetric, all the eigenvalue are also real)

I want to prove that signature of $$b$$ and $$h$$ are the same, however I lack a bit of knowledge in linear algebra and do not know to to prove it?

Let $$\{e_i\}$$ be the real basis of $$V$$ , then it will be the complex basis for $$V_{\Bbb{C}}$$, therefore the matrix associated to $$b$$ has the form $$[b(e_i,e_j)]$$ while the matrix associated to $$h$$ has the form $$[h(e_i,e_j)] = [b(e_i, \bar{e_j})] = [b(e_i, e_j)] = b_{ij}$$ therefore they are the same matrix ?

where $$\bar{e_i} = \overline{e_i\otimes 1} = e_i\otimes \bar{1} = e_i$$?

• Just diagonalize. Sep 8, 2022 at 4:53
• Thank you, Can you have a look at my proof? I found the matrix for both form are the same? It seems a bit weird since the conjugation of the real basis it's again itself. @Qiaochu Yuan Sep 8, 2022 at 5:15
• That's what it means to be a real vector. There's no issue here. Sep 8, 2022 at 5:17
• yeah real is defined to be conjugation invariant. Sep 8, 2022 at 12:38

Given any basis we have the matrix associated to the Hermitian form, it's $$[h_{ij}] = [h(e_i,e_j)] = [b(e_i,\bar{e_j})] = b_{ij}$$ which is real symmetric hence also Hermitian symmetric.
$$h(x,y) = Y^*AX\\b(x,y) = Y^tAX$$