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I learned now about how to diagonalize a symmetric matrices using congruence

for example I saw that there exists an invertible matrix $P$ so that

$P\begin{pmatrix} 3 & 0 & 6\\ 0 & 0 & 1\\ 6 & 1 & 1\\ \end{pmatrix}P^{t}=\begin{pmatrix} 3 & 0 & 0\\ 0 & -11 & 0\\ 0 & 0 & (1/11) \\ \end{pmatrix}$

But how can I diagonalize this symmetric matrix using congruence

$\begin{pmatrix} 0 & 1\\ 1 & 0\\ \end{pmatrix}$

How can I find this $P$ according to the algorithm.

I would have link the algorithm if I would find it in the Net but I am sure you know what I am talking about.

Thanks in advanced !!

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  • In order to diagonalize this matrix, you would need a $P$ with entries that are complex numbers. – Ben Grossmann Jul 15 '13 at 16:45
  • @Omnomnomnom Why? It's a real symmetric matrix. $P = \frac{\sqrt{2}}{2} \begin{bmatrix} 1 & 1 \ -1 & 1 \end{bmatrix}$ should do the trick. – Vedran Šego Jul 15 '13 at 17:03
  • @Omnomnomnom Why? The OP is talking about diagonalisation by congruence, not diagonalisation by similarity. That is, he/she is trying to diagonalise a symmetric bilinear form but not a linear operator. As long as the characteristic of the ground field is not $2$, this is always possible over the same field, not to mention that the matrix in question is real symmetric, so that even the usual orthogonal diagonalisation works. – user1551 Jul 15 '13 at 17:05
  • Ah whoops. Didn't read carefully. – Ben Grossmann Jul 15 '13 at 17:11

1 Answers1

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Hint: You may first borrow some nonzero numbers from other off-diagonal entries. For example, see this answer to another question.

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