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Let $B\bf $ be a real symmetric matrix of order $n\times n$. Then show that there exists an invertible matrix $\bf P$ such that $\bf P'AP=\begin{pmatrix}\bf I & \bf0 & \bf0 \\\bf0 & \bf-I & \bf0 \\\bf0 & \bf0 & \bf0* \\\end{pmatrix}$; $\bf I$ and $\bf-I$ denote the number of 1s and -1s in the diagonal and $\bf0*$ denotes the number of 0s.

I seriously have no idea now to begin this proof and cannot find any resources online which state the proof.

Can anyone help me out or redirect me to the proof?

Abdul Miah
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  • $I$ and $-I$ are matrices. They cannot denote numbers. And you might google on “inertia” of symmetric matrices. – Chris Godsil Nov 19 '17 at 23:27

1 Answers1

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This is a basic result called Sylvester's law of inertia. You can find the proof in any good book on Linear Algebra. See for instance

J. J. Rotman, Advanced Modern Algebra (second edition), Theorem 8.84 p. 692.

Francesco Polizzi
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  • algorithm that gets as far as diagonal, after which one may use a final diagonal $P_{k+1}$ (with some square roots) to reach $\pm 1.$ Did this one yesterday: https://math.stackexchange.com/questions/2495163/quadratic-forms-completing-squares/2495265#2495265 – Will Jagy Oct 30 '17 at 17:31
  • also first edition Rotman, page 704 Theorem 9.83 – Will Jagy Oct 30 '17 at 17:37