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From: HardySpicer on 7 May 2010 17:35
For a real symmetric matrix R, I define its square root as


R=A^0.5 X (A^0.5)'

where ' is transpose. Now I can put U'U =I where U is a unitary matrix
and I is the identity matrix and squeeze that in-between so that the
solution is not unique.


ie
R=A^0.5 XU'U X (A^0.5)'

= B^0.5 X (B^0.5)'

where B=AU

For a given order of matrix - say n - how many unitary matrices and
hence solutions B exist for this definition of square root?


Hardy
From: Robert Israel on 7 May 2010 18:38
HardySpicer <gyansorova(a)gmail.com> writes:

> For a real symmetric matrix R, I define its square root as
>
>
> R=A^0.5 X (A^0.5)'

I was confused by that X, but I think I see what you mean: you just
want R = P P' where P for some matrix P. Of course (assuming P is real)
this will only work if R is positive semidefinite.

> where ' is transpose. Now I can put U'U =I where U is a unitary matrix
> and I is the identity matrix and squeeze that in-between so that the
> solution is not unique.
>
>
> ie
> R=A^0.5 XU'U X (A^0.5)'
>
> = B^0.5 X (B^0.5)'
>
> where B=AU

I think you mean if R = P P' and U is any orthogonal matrix,
R = P U U' P' = (PU) (PU)'. Conversely, if P P' = Q Q' and
P is invertible, then we can write Q = P U where U = P^(-1) Q
is orthogonal.

> For a given order of matrix - say n - how many unitary matrices and
> hence solutions B exist for this definition of square root?

How many n x n orthogonal matrices are there? Infinitely many, of course,
as long as n > 1. O(n) is a Lie group of dimension n(n-1)/2.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: I.M. Soloveichik on 7 May 2010 15:37
If you diagonalize the matrix, then taking square roots of each diagonal entry you would expect 2^n different square roots of a matrix of size n.

> For a real symmetric matrix R, I define its square
> root as
>
>
> R=A^0.5 X (A^0.5)'
>
> where ' is transpose. Now I can put U'U =I where U is
> a unitary matrix
> and I is the identity matrix and squeeze that
> in-between so that the
> solution is not unique.
>
>
> ie
> R=A^0.5 XU'U X (A^0.5)'
>
> = B^0.5 X (B^0.5)'
>
> where B=AU
>
> For a given order of matrix - say n - how many
> y unitary matrices and
> hence solutions B exist for this definition of
> square root?
>
>
> Hardy
From: fernando revilla on 8 May 2010 02:12
I.M. Soloveichik wrote:

> If you diagonalize the matrix, then taking square
> roots of each diagonal entry you would expect 2^n
> different square roots of a matrix of size n.

However, fixing previously f: D C K -> K (K = R or K = C ),
for every nxn matrix A with minimal polynomial:

mu ( x ) = (x - a_1)^{m_1} ... (x - a_s)^{m_s}

such that exist:

f( a_1 ) , ... , f^{ ( m_1-1 )}( a_1 )
...
f( a_s ) , ... , f^{ ( m_s-1 ) }( a_s )

the existence and uniqueness of f( A ) is guaranteed by
means of f (A ) = : p (A ), where p is the corresponding
Lagrange-Sylvester polynomial.

---
http://ficus.pntic.mec.es/~frej0002/
 | 
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