how to solve B = X'AX for matrix X?

[David Fass]

Hi. Perhaps this is a silly question, but how do you solve the
matrix equation X'AX = B for square matrix X?

Thanks for any tips.

-- Dave

[Richard Hindmarsh]

David Fass wrote:
>
>
> Hi. Perhaps this is a silly question, but how do you solve the
> matrix equation X'AX = B for square matrix X?
>

This may be a Riccati eqn. - if you have the Control Toolbox try
CARE, it may work but depends on the properties of A and B, in
particular B has to be symmetric.

In general, an iteration is need, but whether this iteration
converges depends again on the properties of A and B. No simple
answer I'm afraid

no numbers in e-mail

[Peter Spellucci]

In article <eecc2...@webx.raydaftYaTP>,

dpends on A and B .
here is a solution from nanet:
\begin{ citation}:

Given two symmetric positive definite matrices
A = A' > O and H = H' > O
of the same dimensions, we seek another such X = X' > O satisfying the
quadratic equation
XAX = H .

To construct X , first factor A = U'U and let R be the inverse of U.
An acceptable choice for U is the symmetric positive definite square
root of A . Another choice, better numerically, is the upper-triangular
factor U obtained by Choleski's method, and then its inverse R need
not be computed explicitly. Anyway, however U may be found, then let
V = V' > O be the symmetric positive definite square root of UHU' = V^2 .
This V can be computed from an eigendecomposition of UHU' .
Finally X := RVR' .

Despite the variety of choices for U , there is only one symmetric
positive definite solution X of the given equation XAX = H .

W. Kahan
\end{citation}
there is another solution, more formally:
\begin{citation}
The SPD solution to the Ricatti equation XAX = B is given by
the "geometric mean" of the matrices A and B. It is given
explicitely by any of the equivalent six expressions:

X = A(A^{-1} B)^{1/2}
= B(B^{-1} A)^{1/2}
= (B A^{-1})^{1/2} A
= (A B^{-1})^{1/2} B
= A^{1/2} (A^{-1/2} B A^{-1/2})^{1/2} A^{1/2}
= B^{1/2} (B^{-1/2} A B^{-1/2})^{1/2} B^{1/2}

When A and B commute, it reduces to (A B)^{1/2}.

Here are some references to the geometric mean of SPD matrices:

W. N. Anderson and G.E. Trapp, "Shorted operators", SIAM J. Appl. Math.,
28 (1975), pp. 60--71.

J. D. Lawson and Y.Lim, "The geometric mean, matrices, metrics and
more", Amer. Math. Monthly, 108 (2001), pp. 797--812.
\end{citation}
hth
peter

[David Fass]

Thanks very much to both of you!

-- Dave

Peter Spellucci wrote:
>
>
>
> In article <eecc2...@webx.raydaftYaTP>,
> "David Fass" <david...@hotmail.com> writes: