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 Algebraic Riccati Equation
 Solving ARE
 Bounded Real Lemma
 Standard ARE

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 Algebraic Riccati Equation:
 A*X+XA+XRX+Q=0, R=R*, Q=Q*
 An ARE may have many solutions. We are only interested in symmetric
solutions. In particular, we are interested in the symmetric solution
such that A+RX is stable. This solution is called stabilizing solution.
 Consider the associated Hamiltonian matrix:
 Then
 so H and H* are similar. Thus l is an eigenvalue iff 
l* is.
 Thus we conclude: eig(H) ¹ jw Û H has n
eigenvalues in Re s<0 and n eigenvalues in Re s>0.

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 Let X_(H) be the ndimensional spectral subspace corresponding to
eigenvalues in Re s <0:
 where X_{1}, X_{2}
Î C^{n}^{´}^{n}
(X_{1} and X_{2} can be chosen to be real
matrices.) If X_{1}
is nonsingular, define
 X:=Ric(H)= X_{2} X_{1}^{1}: dom(Ric) Ì R^{2n}^{´}^{2n}
® R^{2n}^{´}^{2n}
 where dom(Ric) consists of all H matrices such that
 H has no eigenvalues on the imaginary axis
 are
complementary (or X_{1} is nonsingular.)
 Then X is a solution of the ARE. (see next theorem)
 >>[X_{1},X_{2}] = ric_schr(H), X = X_{2}/X_{1}

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 Theorem 12.1: Suppose H Î dom(Ric) and X=Ric(H). Then
 (i) X is real symmetric
 (ii) X satisfies the algebraic Riccati equation
 A*X+XA+XRX+Q=0
 (iii) A+RX is stable.
 Proof: (i) Let . We show X_{1}*X_{2} is
symmetric.
 Note that there exists a stable matrix H_ in R^{n}^{´}^{n}
such that
 Premultiply this equation by
 to get

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 Since JH is symmetric Þ :
 (X_{1}*X_{2}+X_{2}*X_{1}) H_=H_*(X_{1}*X_{2}+X_{2}*X_{1}
)*= H_*(X_{1}*X_{2}+X_{2}*X_{1} )
 This is Lyapunov equation. Since H_ is stable, the unique solution is
 X_{1}*X_{2}+X_{2}*X_{1 }=0.
 i.e., X_{1}*X_{2} is symmetric. Þ X=(X_{1}^{1})*(X_{1}*X_{2})X_{1}^{1}
is symmetric.
 (ii) Start with the equation
 and postmultiply by X_{1}^{1} to get
 now premultiply by [X I]:
 This is precisely the Riccati equation.
 (iii)
 Thus A + RX is stable because H_ is.

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 Theorem 12.2: Suppose eig(H) ¹ jw and R is semidefinite (³ 0 or £0). Then H Î dom(Ric) Û (A, R)
is stabilizable.
 Proof. (Ü ) Note
that
 We need to show that X_{1} is nonsingular, i.e., Ker X_{1}=0.
 Claim: Ker X_{1} is H_invariant.
 Let x Î Ker X_{1}
and note that X_{2}*X_{1} is symmetric and
 AX_{1}+RX_{2}=X_{1}H_
.
 Premultiply by x* X_{2}* , post multiply by x to get
 x* X_{2}* RX_{2} x=0 Þ RX_{2}
x=0 Þ X_{1}H_ x=0
 i.e., H_ x Î Ker
X_{1}.
 Suppose Ker X_{1}¹0. Then H_ _{Ker}_{ X}_{1}
has an eigenvalue, l, and a corresponding eigenvector, x:
 H_ x= l x, Rel <0, 0¹ xÎ Ker X_{1}

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 Note that
 QX_{1}A*X_{2}=X_{2 }H_
 Postmultiply the above equation by x:
 (A*+lI)X_{2} x=0
 Recall that RX_{2}x = 0, we have
 x*X_{2}*[A+ l*I R]=0.
 (A,R) stabilizable Þ X_{2} x=0 Þ
 has full column rank, which is a
contradiction.
 (Þ ) H Î dom(Ric) Þ A+RX is stable Þ (A, R) is
stabilizable.

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 Corollary 12.3: Let g>0. G(s)=C(sIA)^{1}B+D Î RH_{¥} and
 where R=g^{2}ID*D.
Then the following conditions are equivalent:
 (i) G_{¥}_{ }< g .
 (ii) D< g and H has no eigenvalues on the
imaginary axis.
 (iii) D< g and H Î dom(Ric).
 (iv) D< g
, H Î dom(Ric)
and Ric(H)³0
(Ric(H)>0 if (C,A) is observable).

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 (v) D< g and
there exists an X³0 such that
 X(A+BR^{1}D*C)+(A+BR^{1}D*C)*X+XBR^{1}B*X+C*(I+DR^{1}D*)C=0
 and A+BR^{1}D*C+BR^{1}B*X has no eigenvalues on the imaginary
axis.
 (vi) D< g and
there exists an X > 0 such that
 X(A+BR^{1}D*C)+(A+BR^{1}D*C)*X+XBR^{1}B*X+C*(I+DR^{1}D*)C<0
 (vii) there exists and X > 0 such that
 Proof: We have already known: (i) Û (ii).
(iii) Þ (ii) is obvious. To show that (ii) Þ (iii), we need to show that (A+BR^{1}D*C,
BR^{1}B*) is stabilizable (Theorem 12.2). In fact, we will show
that A+BR^{1}D*C is
stable for all those g such that G_{¥}_{ }< g .

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 Note that we can write
 A+BR^{1}D*C= A+B(g^{2}ID*D)^{1}D*C= A+B(IDD_{1})^{1}DC_{1}
 with D=D*/g, D_{1 }=D/g, and C_{1 }=C/g. Then D<1 and
 C_{1}(sIA)^{1}B+D_{1}_{¥}_{ }= g^{1} G_{¥}_{ } <1.
 Hence by small gain theorem, A+B(IDD_{1})^{1}DC_{1 } is stable for all D with D<1. Thus A+BR^{1}D*C
is stable for all g^{} such that G_{¥}_{ } <g^{}.
 (iii) Þ (iv) follows from the fact that the
ARE
 X(A+BR^{1}D*C)+(A+BR^{1}D*C)*X+XBR^{1}B*X+C*(I+DR^{1}D*)C=0
 can be regarded as a Lyapunov equation with
 A_{1} := A+BR^{1}D*C, Q := XBR^{1}B*X+C*(I+DR^{1}D*)C
 Hence X ³ 0 since
A_{1} is stable and Q³0.
 (v) Þ (i): Assume D = 0 for simplicity. Then
there is an X³0
 and A+BB*X/g^{2} has no jwaxis eigenvalue.

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 Hence
 has no zeros on the imaginary
axis since
 has no poles on the imaginary axis. Next, note that
 X(jwIA)(jwIA)*X+XBB*X/g^{2}I +C*C=0
 Multiply B*{(jwIA)*}^{1}
on the left and (jwIA)^{1}B on the right of the above
equation to get
 B*{(jwIA)*}^{1}XBB*X(jwIA)^{1}B+B*{(jwIA)*}^{1}C*C
(jwIA)^{1}B
 + B*{(jwIA)*}^{1}XBB*X(jwIA)^{1}B/g^{2} =0
 Completing square, we have
 G*(jw)G(jw)= g^{2}IW*(jw)W(jw)

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 Since W(s) has no jwaxis zeros, we conclude that G_{¥}_{ }<
g .
 (vi) Û (vii): follows from Schur complement.
 (vi) Þ (i): by following the similar
procedure as above.
 (i) Þ (vi): let
 Then there exists an e>0 such that G_{e}_{¥}_{ }<
g . Now
(vi) follows by applying part (v) to G_{e}_{ }.

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 Theorem 12.4: Suppose H has the form
 Then H Î dom(Ric)
iff (A,B) is stabilizable and (C,A) has no unobservable modes on the
imaginary axis. Furthermore, X=Ric(H)³0. And X > 0 if and only if (C,A) has no
stable unobservable modes.
 Proof: Only need to show that, assuming (A,B) is stabilizable, H has no jw eigenvalues iff (C,A) has no
unobservable modes on the imaginary axis.
 Suppose that jw
is an eigenvalue and
is a corresponding eigenvector. Then
 AxBB*z=jw x, C*CxA*z=jw z
 Rearrange: (A jwI)x=BB*z, (AjwI)*z=C*Cx

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 Thus <z, (A jwI)x>=<z,BB*z>=B*z^{2}
  <x, (A jwI)*z>=<x,C*Cx>=Cx^{2}
 so <x,(AjwI)*z>
is real and
 Cx^{2}=< (AjwI)x,z>=<z,(AjwI)x>*=B*z^{2}
 Therefore B^{*}z = 0 and Cx = 0. So
 (AjwI)x=0, (AjwI)*z=0
 Combine the last four equations to get
 The stabilizability of (A,B) gives z=0. Now it is clear that jw is an eigenvalue of H
if jw is an
unobservable mode of (C,A).
 (ABB*X)*X+X(ABB*X)+XBB*X+C*C=0
 X³0 since ABB^{*}X
is stable.

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 Corollary 12.5: Suppose (A,B) is stabilizable and (C,A) is detectable.
Then
 A*X+XAXBB*X+C*C=0
 has a unique positive semidefinite solution. Moreover, it is
stabilizing.
 Corollary 12.7: Suppose D has full column rank and denote R = D^{*}D>0.
Let H have the form
 Then H Î dom(Ric)
iff (A,B) is stabilizable and has full column
rank for all w. Furthermore, X=Ric(H)³0 if H Î dom(Ric) and Ker(X) = 0 if and only if (D_{^}*C, ABR^{1}D*C)
has no stable unobservable modes.
 Proof: This is because has full
column rank for all w
 Û ((IDR^{1}D*)C,
ABR^{1}D*C) has no observable modes on jwaxis.
