Solution of Quadratic Matrix Equation

Hello!
I want to solve a Quadratic Matrix Equation of the form -
Solve for z: z' A z + B' z + D = 0
I did see a similar question here and David Goodmanson has an excellent answer for the z' C z + z + D = 0. However, I don't understand how he came up with the closed form solution or how to generalize it for my case. Any help would be greatly appreciated.

7 Commenti

What are the dimensions of the scalars/vectors/matrices involved ?
Neilabh Banzal
Neilabh Banzal il 25 Feb 2022
Modificato: Neilabh Banzal il 25 Feb 2022
@Torsten, A is 6x6x6, B is 6x6, D is 6x1, z is 6x1.
You will have to tell us how to multiply A with z.
Neilabh Banzal
Neilabh Banzal il 25 Feb 2022
Modificato: Neilabh Banzal il 25 Feb 2022
@Torsten, I can compute Az as a function of z which is 6x6.
Then you have 6 equations for 6 unknowns.
Use fsolve to solve for z1,...,z6.
I was trying to avoid using fsolve as I have to call it repeatedly. Is there no other way?
Torsten
Torsten il 25 Feb 2022
Modificato: Torsten il 25 Feb 2022
Maybe - if you tell us how you plan to evaluate A*z to get a 6x6-matrix.

Accedi per commentare.

Risposte (2)

Benjamin Thompson
Benjamin Thompson il 24 Feb 2022
icare and idare are used to numerically solve the continuous or discrete forms of the Riccati equation.

1 Commento

@Benjamin Thompson, this is not a Riccati Equation. Note that there is a B' z term, but no z' B term.

Accedi per commentare.

Benjamin Thompson
Benjamin Thompson il 25 Feb 2022
You could look at this work on File Exchange:
Then you run it 6 times for each of the 6 6x6 matrices in the 6x6x6 A matrix. If you have specific values to post for your problem in a MAT file, please do so.

2 Commenti

But according to the OP, z is 6x1, not 6x6.
Ok, I don't understand his problem requirements either but hopefully this helps him out.

Accedi per commentare.

Prodotti

Release

R2021b

Richiesto:

il 24 Feb 2022

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by