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G is a positive-definite symmetric matrix, V is another symmetric matrix,ie $V\neq G$, All their elements are Real number. How to find a matrix O such that:

$$OGO^{T}=I$$ $$OVO^{T}=Diag(\lambda_1,\lambda_2...\lambda_n)$$

xjtein
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  • Such a matrix may not exist. – 5xum Sep 20 '15 at 14:50
  • In one paper, it says that such matrix always exist. – xjtein Sep 20 '15 at 14:54
  • Are $\lambda_1,\dots$ eigenvalues of $V$? – 5xum Sep 20 '15 at 14:55
  • I think so but not necessary. All I want is make V diagonal. – xjtein Sep 20 '15 at 14:59
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    Since the case $V=G$ is allowed by the conditions specified in the question there is clearly a problem somwhere. – Shahar Even-Dar Mandel Sep 20 '15 at 15:00
  • You think so? Sorry to say this but if you don't even put in the effort to post a question you understand, don't expect good answers to your question. – 5xum Sep 20 '15 at 15:00
  • Sorry, what I mean is that I not very sure, all I want to say is that there exist O to preserve G and diagonize V. By preserving G, we can get det(O)=1, so definitely Product of all $\lambda$ s are the product of eigenvalues of V, but I can not prove that they are the same. – xjtein Sep 20 '15 at 15:07
  • Just a remark: if you replace $OGO^T=G$ with $OGO^T=I$ you will get the simultaneous diagonalization of two Hermitian matrices. I suspect your problem is somehow related, but you are missing an extra relation between $V$ and $G$. – A.Γ. Sep 20 '15 at 15:15
  • xjtan, *you need to learn to give the actual reference, title and authors, also typeset the actual equation as part of your original question,* and explain why you think that equation becomes the simplified matrix question you are asking here. – Will Jagy Sep 20 '15 at 17:02
  • Will, thanks for your advice on how to ask questions. – xjtein Sep 21 '15 at 01:09
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    After I take Will Jagy's advice, I try to edit the question and go back to the original paper I was reading, and find that I misunderstood the paper and asked an wrong question. The author means simultaneous diagonalization of two Hermitian matrices just as A.G. mentioned. I am so terribly sorry to waste all you guys' time. I will be much more careful in the future. Sorry guys. – xjtein Sep 21 '15 at 01:52

1 Answers1

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For sake of having an answer, this is called simultaneous diagonalisation by congruence and the method is well-known. Since $G$ is positive definite, it has a unique positive definite square root $G^{1/2}$. Now let $O=QG^{-1/2}$ for some real orthogonal matrix $Q$. Then the first equation is automatically satisfied, and the second one means that $QG^{-1/2}VG^{-1/2}Q^T$ is diagonal. Now the latter is solvable because $G^{-1/2}VG^{-1/2}$ is real symmetric and hence real orthogonally diagonalisable.

user1551
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