17.4 End Game

Now let us compare what we have found to the well known solution of (HGRAIL). In that theory there are two Riccati equations due to Doyle, Glover, Khargonekar and Francis. These are the DGKF X and Y equations. One can read off that the E11 equation which we found is the DGKF equation for Y -1, while the Riccati equation which we just analyzed is the DGKF X equation.

Indeed what we have proved is that if (HGRAIL) has a solution with Eij invertible and if b and c are given by formulas Out[8] and Out[9] in §17.3.2, then

(1) the DGKF X and Y -1 equations must have a solution
(2) X and Y are self-adjoint
(3) Y -1 - X is invertible

Now we turn to the converse. The straightforward converse of the above italicized statement would be: If items (1), (2) and (3) above hold, then (HGRAIL) has a solution with Eij invertible and b and c are given by formulas Out[8] and Out[9] in §17.3.2. There is no reason to believe (and it is not the case) that b and c must be given by the formulas Out[8] and Out[9] in §17.3.2. These two formulas came about in §17.3.2 and were motivated by “excess freedom” in the problem. The converse which we will attempt to prove is:

Proposed Converse 17.1 If items (1), (2) and (3) above hold, then (HGRAIL) has a solution with Eij invertible.

To obtain this proposed converse, we need a complete spreadsheet corresponding to the last stages of our analysis. The complete spreadsheet is:

THE ORDER IS NOW THE FOLLOWING: A < AT < B1 < B1T < B2 < B2T < C1 < C1T < C2 < C2T < X < X-1 < Y < Y -1 E12 E21 E22 E12T E21T E22T E11 E11T E11-1 E11-1T E12-1 E21-1 E22-1 E12-1T E21-1T E22-1T b bT c cT a aT _________________________________________________________________________________ __________________________  YOUR SESSION HAS DIGESTED _____________________________________ __________________________  THE FOLLOWING RELATIONS ______________________________________ _________________________________________________________________________________ THE FOLLOWING VARIABLES HAVE BEEN SOLVED FOR:
{a,b,c,E11,E11-1,aT ,bT ,cT ,E11T ,E12T ,E21T ,E22T ,E11-1T ,E12-1T ,E21-1T ,E22-1T }
The corresponding rules are the following:

a →-E12-1 AT E12+E12-1 C1T B2T E12+E12-1 C2T B1T E12+E12-1 E11 B2 B2T E12- E12-1 C1T C1 E21-1 E22 -E12-1 E11 B2 C1 E21-1 E22 -E12-1 C1T B2T E11 E21-1 E22 - E12-1 E11 B2 B2T E11 E21-1 E22

b →-E12-1 C2T - E12-1 E11 B1

c →-B2T E12 + C1 E21-1 E22 + B2T E11 E21-1 E22

E11 Y -1

E11-1 Y

aT → -E21 AE21-1 + E21 B1 C2 E21-1 + E21 B2 C1 E21-1 + E21 B2 B2T E11 E21-1 - E22 E12-1 C1T C1 E21-1 -E22 E12-1 E11 B2 C1 E21-1 -E22 E12-1 C1T B2T E11 E21-1 - E22 E12-1 E11 B2 B2T E11 E21-1

bT →-C2 E21-1 - B1T E11 E21-1

cT →-E21 B2 + E22 E12-1 C1T + E22 E12-1 E11 B2

ET  → E11           ET  → E21           ET  → E12           ET  → E22
 1-11T     -1         1-21T     -1         2-11T     -1         2-21T     -1
E11  →  E 11         E12  →  E 21         E21  →  E 12         E22  →  E 22
_________________________________________________________________________________
The expressions with unknown variables {}
and knowns {A,B1,B2,C1,C2,X,Y,X-1,Y -1,AT ,B1T ,B2T ,C1T ,C2T }
X X-1 1

Y Y -1 1

X-1 X 1

Y -1 Y 1

Y -1 B1 C2 Y -1 A + AT Y -1 + C1T C1 - C2T C2 - C2T B1T Y -1

X B2 B2T X X A + AT X - X B2 C1 - C1T B2T X + X B1 B1T X

_________________________________________________________________________________ ________________________  USER CREATIONS APPEAR BELOW ___________________________________ _________________________________________________________________________________
E11-1 Y

E12 E22-1 E21 E11 - X

_________________________________________________________________________________ ____________________  SOME RELATIONS WHICH APPEAR BELOW _____________________________ ______________________________  MAY BE UNDIGESTED ___________________________________________ _________________________________________________________________________________ THE FOLLOWING VARIABLES HAVE NOT BEEN SOLVED FOR: {E12,E21,E22,E12-1,E21-1,E22-1}
_________________________________________________________________________________
1.3 The expressions with unknown variables {E12-1,E12}
and knowns {}
E12 E12-1 1

E12-1 E12 1

_________________________________________________________________________________
1.3 The expressions with unknown variables {E21-1,E21}
and knowns {}
E21 E21-1 1

E21-1 E21 1

_________________________________________________________________________________
1.3 The expressions with unknown variables {E22-1,E22}
and knowns {}
E22 E22-1 1

E22-1 E22 1

_________________________________________________________________________________
4.3 The expressions with unknown variables {E22-1,E11,E21,E12}
and knowns {X}
 E12 E22-1 E21 E11 - X

In the spreadsheet, we use conventional X, Y -1 notation rather than “discovered” notation so that our arguments will be familiar to experts in the field of control theory.

Now we use the above spreadsheet to verify the proposed converse. To do this, we assume that matrices A, B1, B2, C1, C2, X and Y exist, that X and Y are invertible, that X and Y are self-adjoint, that Y -1 - X is invertible and that the DGKF X and Y -1 equations hold. That is, the two following polynomial equations hold.

Y -1 B 1 C2 = Y -1 A + AT Y -1 + C 1T C 1 - C2T C 2 - C2T B 1T Y -1
X B2 B2T X = X A + AT X - X B 2 C1 - C1T B 2T X + X B 1 B1T X

We wish to assign values for the matrices E12, E21, E22, E11, a, b and c such that each of the equations on the above spreadsheet hold. If we can do this, then each of the equations from the starting polynomial equations from §17.1 will hold and the proposed converse will follow.

(1) Note that all of the equations in the {}-Category of the above spreadsheet hold since X and Y solve the DGKF equations and are both invertible.
(2) Set E11 equal to the inverse of Y . This assignment is dictated by the user selects. Note that E11 = E11T follows since Y is self-adjoint.
(3) Let E12 and E21 be any invertible matrices such that E12T = E 21. For example, one could choose E12 and E21 to both be the identity matrix.
(4) Note that there is there is a user select E12 E22-1 E 21 = E11 - X and that E12, E21 are invertible. Since Y -1 - X is invertible and E 11 = Y -1, E 11 - X are invertible. Therefore, we set E22 = E12-1(E 11 - X)-1E 21-1. Since E 12T = E 21, E11T = E 11 and XT = X, it follows that E 22 is invertible and self-adjoint.
(5) Since Eij has been set for i,j = 1, 2, we can set a, b and c according to their formulas at the top of the spreadsheet .

With the assignments of E12, E21, E22, E11, a, b and c as above, it is easy to verify by inspection that every polynomial equation on the spreadsheet above holds.

We have proven the proposed converse and, therefore, have proven the following approximation to the classical [DGKF] theorem.

Theorem 17.2 If (HGRAIL) has a solution with invertible Eij and b and c are given by the formulas Out[8] and Out[9] in §17.3.2, then the DGKF X and Y -1 equations have solutions X and Y which are symmetric matrices with X,Y -1 and Y -1 - X invertible. The DGKF X and Y -1 equations have solutions X and Y which are symmetric matrices with X,Y -1 and Y -1 - X invertible, then (HGRAIL) has a solution with invertible E ij.

Note that we obtained this result with an equation in the one unknown X and an equation with the one unknown E11 = Y -1. From the strategy point of view, the first spreadsheet featured an equation in the single unknown b (and its transpose) and an equation in the single unknown c (and its transpose) and so is the most complicated. For example, the polynomial Out[4] in §17.3.2 decomposes as

p = qT1 q1 + q2
(17.3)

where q1 = b + E12-1 C 2T + E 12-1 E 11 B1 and q2 is a symmetric polynomial which does not involve b. This forces us to say that the proof of the necessary side of Theorem 17.2 was done with a 2-strategy.

A more aggressive way of selecting knowns and unknowns allows us to obtain this same result with a symmetrized 1-strategy. In particular, one would set a, b and c to be the only unknowns to obtain a first spreadsheet. The first spreadsheet contains key equations like (17.3), which is a symmetric 1-decomposition, because q2 does not contain a, b or c. Once we have solved for a, b and c, we turn to the next spreadsheet by declaring the variables involving Eij (e.g., E11, E11-1, . . . ) to be unknown. At this point, the computer run is the same as Steps 2, 3 and 4 above.