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Can we say any matrix whose quadratic form is $xy + z^2$ can be diagonalised to a Matrix who has two positive and one negative Eigen value?

If it can , from which theorem you can. I have been having a really bad time to understand one problem.

I would be highly obliged if anyone help me to get my doubt cleared.

Thank You in advance.

cmi
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1 Answers1

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Yes it is possible by Sylvester's law of inertia since the matrix associated to $xy+z^2$ is

$$B=\begin{bmatrix}0&\frac12&0&0\\\frac12&0&0&0\\0&0&1&0\\0&0&0&0 \end{bmatrix}$$

with signature

$$n_+=2,n_-=1,n_0=1$$

user
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  • Well the part of signature zero is totally useless. – Riccardo Ceccon Apr 13 '18 at 06:36
  • @RiccardoCeccon Each one is useless with respect to the others since the sume is equal to n. I like give all of them. – user Apr 13 '18 at 06:44
  • $$B=\begin{bmatrix}0&\frac13&0&0\\frac23&0&0&0\0&0&1&0\0&0&0&0 \end{bmatrix}$$...This is also a matrix which satisfies that quadratic form. There may be thousands other. How can you ensure that there can not be any other matrix whose diagonal elements are some thing different than what is said , satisfies $xy + z^2$. You have been sticking to a single Matrix since yesterday.@gimusi – cmi Apr 13 '18 at 06:45
  • how the associated matrix ensuring that any matrix which satisfies $xy +z^2$ has the same rank and signature?@gimusi – cmi Apr 13 '18 at 06:48
  • @cmi This is the symmetric matrice associated to the qudratc form, that is the important one since the skew part turns out to be zero. Refer also to https://math.stackexchange.com/questions/307381/why-do-we-assume-that-a-matrix-in-quadratic-form-is-symmetric – user Apr 13 '18 at 06:49
  • @cmi The unique symmetric part of those matrices can be diagonalized and then for any B we can write $x^TBx=x^TS^TDSx=y^TDy$. And D has the same signature (and rank) of the symmetric part of $B$. – user Apr 13 '18 at 07:00
  • so you are claiming that there is an unique symmetric Matrix for each quadratic form when the order of the matrix is fixed?@gimusi – cmi Apr 13 '18 at 07:01
  • @cmi yes of course, for a given basis. – user Apr 13 '18 at 07:03
  • @cmi Read carefully, I wrote "for a given basis". – user Apr 13 '18 at 07:05
  • yes I am asking you what is the meaning of "for a given basis"? – cmi Apr 13 '18 at 07:06
  • I am on the verge of understanding the problem. I hope you would not leave me curious@gimusi – cmi Apr 13 '18 at 07:10
  • sorry....yes I am asking you what is the meaning of "for a given basis"? – cmi Apr 13 '18 at 07:12
  • @cmi Let consider the QF $x^TBx$ with respect to a given basis (for example the standard one). Let M be the matrix of the change of basis that is $x=My$ then $x^TBx=y^TM^TBMy=y^TCy$ and $C$ is the matrix which represent the same QF in the new basis. – user Apr 13 '18 at 07:13
  • I got you...Now let me take some time to sum up the whole thing.@gimusi – cmi Apr 13 '18 at 07:15
  • @cmi It was an hard task form yesterday! But well done, if you can get it I'm very happy for that! Do not hesitate to ask for any other doubts! See you later. Bye – user Apr 13 '18 at 07:17
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    @cmi I suppose integration in $R^n$, let make a proper OP on that and of course you will obtain an answer – user Apr 17 '18 at 16:51