Let T=[A B;C D] denote a (m+n)X(m+n) real matrix. Let V denote a orthogonal matrix such that V^{T}TV=[M_{1,1} 0;M_{2,1} M_{2,2}]. Partition V=[V_{1,1} V_{1,2};V_{2,1} V_{2,2}] where the main blocks has sizes mXm and nXn.
If V_{2,2} is invertible then one can show that X=-V_{1,2}V_{2,2}^{-1} is a real solution to the above mentioned nonsymmetric algebraic Riccati equation.
My question is:
When (D,C) is controllable is it true that there always exist orthogonal matrix V as above with invertible V_{2,2} block ?
Or, unrelatedly to orthogonal matrices:
Is it true that the nonsymetric algebraic Riccati equation XCX+XD-AX-B=0 always has a real solutions when (D,C) is controllable ?
Dear Yossi,
I do not think this is true. Take C=I, A=0, D=0. Then the pair (D,C) is controllable, and the Riccati equation becomes X^2=B. This has a solution only if B has a square root, and not every matrix has a square root, and certainly not every real matrix has a real square root. Or am I missing something here?
Dear Proffessor A.C.M. Ran !
It is good to hear from you.
You are right !
I forgot to say that T is assumed to be stable.
In this version, my question is my conjecture from 2012 raised since I ran a lot of examples in my computer and I always (for T stable and (D,C) controllable) had a solution to the Riccati equation.
Unfortunately I jave found a counter example:
A=-I, B=[0 1;0 0], C=I, D=-I.
The eigenvalues of T are {-1,-,1,-1,-1} and (D,C) controllable but there is no solution X for the Riccati equation.
I also came up with a proof that there exist X solving the Riccati equation if and only if there exist orthogonal matrix V as above, with block V_{2,2} invertible.
Relating to these observations my question now is:
Uniformly at random choosing stable matrix T with (D,C) controllable, what is the probability of existence of X solving the Riccati equation ?
Or equivalently, what is the measure of the set of all stable matrices T with (D,C) controllable that has X solving the Riccati equation ?
As I said above, in my computer experience I havent tackeled with T as above that didnt have X as above.
I hope to hear from you soon !
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