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

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  • $\begingroup$ Just diagonalize. $\endgroup$ Sep 8, 2022 at 4:53
  • $\begingroup$ 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 $\endgroup$
    – yi li
    Sep 8, 2022 at 5:15
  • $\begingroup$ That's what it means to be a real vector. There's no issue here. $\endgroup$ Sep 8, 2022 at 5:17
  • $\begingroup$ yeah real is defined to be conjugation invariant. $\endgroup$
    – yi li
    Sep 8, 2022 at 12:38

1 Answer 1

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I found the reference in Artin's Algebra book Chapter 8. This chapter answers my question.

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.

Although the matrix looks the same but the expression are the not same , that is :

$$h(x,y) = Y^*AX\\b(x,y) = Y^tAX$$


As Qiaochu point out we can always diagonalize it, even if the form is degenerate (but needs symmetric , hermitian condition.), which is proved on Artin's book theorem 8.4.10.

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