I want to solve the following matrix equation with respect to the matrix variable X which is a real symmetric positive definite matrix. The given matrix A is real and symmetric, "a" is a scalar, and I is the identity matrix of the appropriate size.
X-1X-1+ (X - I)-1 + aX = A
If we ignore the first term in this equation, there is a closed form solution in the following paper, eq. (14) and Lemma 2.1 (I need a wise closed form solution like this one!)
http://www.optimization-online.org/DB_FILE/2009/09/2409.pdf
Dear Janusz
Thanks for your time and the notes. Yes, that is the straitghforward answer. I have tried it previously but it will not lead to a closed form solution. Actually, this equation is part of a bigger problem and it is required to have a closed form matrix solution for this part (to be able to proceed with the rest), i.e. in the form of X =f(A, a, ...) for some function f.
Thanks again,
Mahmoud
Dear Mahmoud Ramezani-Mayiami
X-1X-1+ (X - I)-1 + aX = A
Multiply from left by XX:
XXX-1X-1+ XX(X - I)-1 + XXaX =XX A
I+ XX(X - I)-1 +aXXX =XX A
Multiply from right by (X - I):
(X - I)+ XX + aXXX(X - I) =XX A(X - I):
XX=(X - I)(AXX-I-aXXX)
Multiply from right by X-1 X-1:
I=(X-I)(A-aX)
I=AX-aX^2-A+aX
(A+a*I)X=aX^2+A+I
My opinion:
This could be probably translated into an algebraic polynomial and as well an eigenvalue problem. I am not good at linear algebra and matrix operations, but I guess my operations could possibly present you some ideas to develop further manipulations you yourself. I will think more again about my derivation above.
Dear Saeb AmirAhmadi Chomachar
Thanks for your time. Up to the line XX=(X - I)(AXX-I-aXXX) everything is OK with me. In this line, actually we should have XX=(AXX-I-aXXX)(X - I) instead. But more important is the rest. I do not understand how you get I=(X-I)(A-aX) from that.
Dear Mahmoud Ramezani-Mayiami
There was a mistake in my previous operations.
XX=(X - I)(AXX-I-aXXX)
Factorize !
XX=(X - I)(A-I-aX)XX
Multiply from right by X^-1 X^-1:
I=(X-I)(A-I-aX)
I=XA-A-X^-1+X^-1X^-1-aX^2+aX
Multiply from right by X:
IX=XAX-AX-I+X^-1-aX^3+aX^2
and therefore:
X=XAX-AX-I+X^-1-aX^3+aX^2
This could be solved algebraically, treating it as a matrix polynomial, while knowing the numerical values of matrix A, and scalar a.
Warning: the operations are valid only for symmetric matrices X and A.
Remarks: There might be still some mistakes in my operations. Please check it and feel free to share with me your viewpoints.
Saeb AmirAhmadi Chomachar
First, as I mentioned last time, by your manipulations, we should have XX=(AXX - I -aXXX)(X - I), not (X - I)(AXX - I - aXXX)
Second, (AXX - I - aXXX) is not factorized by (A - I - aX) XX !
To solve cubic eqs, you may look at https://www.storyofmathematics.com/solving-cubic-equations
Muhammad Ali
Solving matrix equation is different than that of scalar polynomial ones
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