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 Posted: Tue Jul 08, 2008 8:35 pm Post subject: Matrix Question |
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If GRG'=I
where G and R are both n square symmetric matrices and I is the
identity matrix of order n (' is transpose).
Then G in terms of R (R is known) is what?
I assume it is
G=R**(-0.5)
where ** is power of the matrix. I assume there are other solutions
however. This is in analogy to the scalar case where
g^2r=1 hence g=1/sqrt(g)
K. |
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Jannick Asmus
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| Posted: Wed Jul 09, 2008 2:39 am Post subject: Re: Matrix Question |
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On 08.07.2008 22:35, kronecker@yahoo.co.uk wrote:
| Quote: |
If GRG'=I
where G and R are both n square symmetric matrices and I is the
identity matrix of order n (' is transpose).
|
This implies that R and G are non-singular, R is positive definite and R
and G are simultaneously diagonalizable.
| Quote: |
Then G in terms of R (R is known) is what?
I assume it is
G=R**(-0.5)
|
This is one solution. If G is not required to be positive definite as
well, G is not uniquely determined. In general, take - R^(-0.5).
Another way to construct solutions is to reduce to the case that R is
equal to a scalar multiple of I (diagonalize and have a look at
eigenspaces): R = aI (a>0). From here you can choose any diagonal matrix
G with G^2 = R = aI.
| Quote: |
where ** is power of the matrix. I assume there are other solutions
however. This is in analogy to the scalar case where
g^2r=1 hence g=1/sqrt(g)
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.... together with -1/sqrt(r) (assuming that you meant 1/sqrt(r) above).
HTH.
--
Best wishes,
J. |
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Jannick Asmus
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| Posted: Wed Jul 09, 2008 11:04 am Post subject: Re: Matrix Question |
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On 08.07.2008 23:39, Jannick Asmus wrote:
| Quote: |
On 08.07.2008 22:35, kronecker@yahoo.co.uk wrote:
If GRG'=I
where G and R are both n square symmetric matrices and I is the
identity matrix of order n (' is transpose).
This implies that R and G are non-singular, R is positive definite and R
and G are simultaneously diagonalizable.
|
We can make it easier from here: Since R and G are simultaneously
*orthogonally* diagonalizable, we can assume w.l.o.g. that R and G be
diagonal matrices with R > 0. Counting the possibilities for the signs
of the diagonal elements of G yields that there are 2^n solutions for G.
| Quote: |
Then G in terms of R (R is known) is what?
I assume it is
G=R**(-0.5)
This is one solution. If G is not required to be positive definite as
well, G is not uniquely determined. In general, take - R^(-0.5).
Another way to construct solutions is to reduce to the case that R is
equal to a scalar multiple of I (diagonalize and have a look at
eigenspaces): R = aI (a>0). From here you can choose any diagonal matrix
G with G^2 = R = aI.
where ** is power of the matrix. I assume there are other solutions
however. This is in analogy to the scalar case where
g^2r=1 hence g=1/sqrt(g)
... together with -1/sqrt(r) (assuming that you meant 1/sqrt(r) above).
K.
HTH.
|
--
Best wishes,
J. |
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