Dear All:

It is not related directly to dynare, but given there are some many experts on dynamic systems and time series, I may find the answer here.

My question is very simple:

What is the general solution for a quadratic matrix equation in the following form:

XAX’+BX’+XC+D=0.

• means muliplication and ’ means transpose.

I know the solution for a similar case when there is no transpose in the above equation, but not this one.

Thanks.

Hi,
Do I understand you well that A,B,C,D all matrices of type n*n? Can we assume that A is regular?

We can use a simple substitution to obtain an equivalent equation without the X’ term:

We want to solve the equation

(1) X * A * X’ + B * X’ + X * C + D = 0.

Consider the following identity

(2) (X + U) * A * (X + U)’ = X * A * X’ + X * A * U’ + U * A * X’+ U * A * U’.

Plugging (2) into (1) gives us

``````    **(X + U) * A * (X + U)' - X * A * U' - U * A * U' + X * C + D = 0**,
``````

where U = B / A. After substituting for Y = X + U, we obtain

``````    **Y * A * Y' - (Y - U) * A * U' - U * A * U' + (Y-U) * C + D = 0**,
``````

or

``````    **Y * A * Y' + Y * (C - A * U') + (D - U * C) = 0**.
``````

The later can be written as

(3) Y * A * Y’ + Y * E + F = 0,

where U = B / A, E = C - A * U’, and F = D - U * C. The equation (3) is what you mentioned you are able to solve. After obtaining the solution for Y, you substitute back for X = Y - U.

Do you know the general solution for the equation (3)? I know only how to solve it for a symmetric matrix A.

Can I ask you how did you get this problem? Do you need to solve equation (1) for some application problem?

Regards,
Pavel

pakocia - How do you solve eqn 3 for a symmetric A? Do you have a reference?

Thanks

To solve the equation

(1) Y * A * Y’ + Y * E + F = 0

we need to assume each of the matrices A, E, F to be symmetric. Then there exist a symmetric solution Y. For any symmetric matrix R, (1) is equivalent to

(2) (Y+R) * A * (Y+R) - 2 Y * A * R - R * R + Y * E + F = 0.

We want to choose R so that

Y * E = 2 * Y * A * R,

so we choose R = 1/2 * (A\E). The equation (2) can be written as

(3) Z * A * Z = G,

where G = R * R - F. At this point we decompose the matrices G and A. We need to assume that A and G are (symmetric) positive definite matrices. By th decomposition we obtain

A = AA * AA, and G = GG * GG.

Then the sufficient condition for Z to be a solution of (3) is

Z * AA = GG,

Which gives us

Z = GG / AA.

This is the way the problem can be solved. I am not sure if it can be solved with less assumptions.

Pavel