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$If$ $two$ $matrices$ $are$ $congruent$ $then$ $they$ $have$ $the$ $same$ $eigen$ $values$ $??$

I have tried solving it using definition of congruent matrices taking eigen value but didn't come up with a proper solution ... how to solve it ???

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  • $\begingroup$ The closest to a true statement similar to this may be Sylvester's inertia theorem ... $\endgroup$ – Hagen von Eitzen May 7 '16 at 22:13
  • $\begingroup$ Like we can prove Aand B are similar matrices then they have the same eigen values | B-kl | = | A-kI | where k is their eigen value .. is there any similr proof to prove | B-kl | is not equal to | A-kI $\endgroup$ – Shona May 7 '16 at 22:22
  • $\begingroup$ What is Sylvester's inertia theorem ? $\endgroup$ – Shona May 8 '16 at 6:35
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It is wrong $\pmatrix{0&1\cr1&0\cr}$ and $\pmatrix{2&1\cr0&-1/2\cr}$ are congruent but the first one has $1$ and $-1$ as eigen values and the second one has $2$ and $-1/2$ as eigen values.

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  • $\begingroup$ Thank you i was trying to prove it using general values .. is there any general solution ?? $\endgroup$ – Shona May 7 '16 at 22:11
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    $\begingroup$ Even simpler(?) $I$ and $(2I)^TI(2I)=4I$ are congruent and have different eigenvalues $\endgroup$ – Hagen von Eitzen May 7 '16 at 22:12
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    $\begingroup$ well it is a counter example, so your statement is false. And imagine that you find a matrix such as $A$ and $D$ are congruent thanks to the orthogonal matrix $P$ and have the same eigen values and $D$ is diagonal. If you multiply $D$ by $\lambda >0$ then you always have $\lambda D$ and $A$ congruent, so they almost never have the same eigen values. $\endgroup$ – Jennifer May 7 '16 at 22:18
  • $\begingroup$ Yes thats true .. what if D is not diagonal $\endgroup$ – Shona May 7 '16 at 22:27
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    $\begingroup$ I have no simple idea at the moment. But not that all the symetric matrices are congruent to a diagonal matrix qnd the diagonal matrices are dence in the spaces of all the matrices, so it is not a very specific case. $\endgroup$ – Jennifer May 8 '16 at 4:37

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