LyapunovRiccati

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Lyapunov Operator

Let A F nn be given, and dene a linear operator LA : C nn C nn asLA (X ) := A X + XA

Suppose A is diagonalizable (what follows can be generalized even if this isnot possible - the qualitative results still hold). Let { 1, 2, . . . , n} be theeigenvalues of A. In general, these are complex numbers. Let V C nn bethe matrix of eigenvectors for A , so

V = [v1 v2 vn ] C nnis invertible, and for each i

A vi = ivi

Let wi C n be dened so that

V 1 =

wT 1wT 2...

wT n

Since V 1V = I n , we have that

wT i v j = ij :=0 if i = j1 if i = j

For each 1 i, j n deneX ij := viv j C nn

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Eigenvalues Lyapunov Operator

Lemma 63 The set {X ij }i=1 ,.,n ; j =1 ,.,n is a linearly independent set, and this set is a full set of eigenvectors for the linear operator LA. Moreover, the eigenvalues of LA is the set of complex numbers i + j i=1 ,.,n ; j =1 ,.,nProof: Suppose { ij }i=1 ,.,n ; j =1 ,.,n are scalars and

n

j =1

n

i=1 ij X ij = 0nn (7.19)

Premultiply this by wT l and postmultiply by wk

0 = wT ln

j =1

n

i=1 ij X ij wk

=n

j =1

n

i=1 ij wT l viv j wk

=n

j =1

n

i=1 ij li kj

= lk

which holds for any 1 k, l n . Hence, all of the s are 0, and the set{X ij }i=1 ,.,n ; j =1 ,.,n is indeed a linearly independent set. Also, by denitionof LA, we have

LA (X ij ) = A X ij + X ij A= A viv j + viv j A= iviv j + vi j v j= i + j viv j= i + j X ij

So, X ij is indeed an eigenvector of LA with eigenvalue i + j .

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Lyapunov Operator Integral Formula

If A is stable (all eigenvalues have negative real parts) then i + j = 0for all combinations. Hence LA is an invertible linear operator. We canexplicitly solve the equation

LA (X ) = Qfor X using an integral formula.

Theorem 64 Let A F nn be given, and suppose that A is stable. The LAis invertible, and for any Q F nn , the unique X F nn solving LA (X ) =Q is given by

X = 0

eA Qe A d

Proof: For each t 0, deneS (t) := t0 eA Qe A d

Since A is stable, the integrand is made up of decaying exponentials, andS (t) has a well-dened limit as t , which we have already denotedX := lim

tS (t). For any t 0,

A S (t) + S (t)A = t0 A eA Qe A + eA Qe A A d = t0

dd

eA Qe A d

= eA tQe At QTaking limits on both sides gives

A X + XA = Qas desired.

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Lyapunov Operator More

A few other useful facts

Theorem 65 Suppose A, Q F nn are given. Assume that A is stable and Q = Q 0. Then the pair (A, Q ) is observable if and only if

X := 0 eA Qe A d > 0Proof: Suppose not. Then there is an xo F n , xo = 0 such that

Qe At xo = 0

for all t 0. Consequently xoeA tQe At xo = 0 for all t 0. Integrat-ing gives

0 = 0 xoeA tQe At xo dt= xo 0 eA tQe At dt xo= xoXx o

Since xo = 0, X is not positive denite.

Suppose that X is not positive denite (note by virtue of the deni-tion X is at least positive semidenite). Then there exists xo F n ,

xo = 0 such that xoXx o = 0. Using the integral form for X (sinceA is stable by assumption) and the fact tha Q 0 gives 0 Q

12 eA xo d = 0

The integrand is nonnegative and continuous, hence it must be 0for all 0. Hence QeA xo = 0 for all 0. Since xo = 0, thisimplies that ( A, Q ) is not observable.

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Lyapunov Operator More

Theorem 66 Let (A, C ) be detectable and suppose X is any solution to A X + XA = C C (there is no apriori assumption that LA is an invertible operator, hence there could be multiple solutions). Then X 0 if and only if A is stable.

Proof Suppose not, then there is a v Cn , v = 0, C , Re( ) 0with Av = v (and v A = v ). Since (A, C ) is detectable and theeigenvalue in the closed right-half plane Cv = 0. Note that

Cv 2 = v C Cv= v (A X + XA ) v= + v Xv= 2Re ( ) v Xv

Since Cv > 0 and Re( ) 0, we must actually have Re ( ) > 0and v Xv < 0. Hence X is not positive semidenite.

Since A is stable, there is only one solution to the equation A X +XA = C C , moreover it is given by the integral formula,

X = 0 eA C Ce A d which is clearly positive semidenite.

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Lyap Ordering

Finally, some simple results about the ordering of Lyapunov solutions.

Let A be stable, so LA is guaranteed to be invertible. Use L1A (Q) todenote the unique solution X to the equation A X + XA = Q .Lemma 67 If Q1 = Q1 Q2 = Q2, then

L1A (Q1) L1A (Q2)

Lemma 68 If Q1 = Q1 > Q 2 = Q2, then

L1A (

Q1) >

L1A (

Q2)

Proofs Let X 1 and X 2 be the two solutions, obviously

A X 1 + X 1A = Q1A X 2 + X 2A = Q2

A (X 1 X 2) + ( X 1 X 2) A = (Q1 Q2)Since A is stable, the integral formula applies, and

X 1 X 2 = 0 eA (Q1 Q2) eA d Obviously, if Q1 Q2 0, then X 1 X 2 0, which proves Lemma 5. If Q1 Q2 > 0, then the pair ( A, Q 1 Q2) is observable, and by Theorem3, X 1 X 2 > 0, proving Lemma 6.

Warning: The converses of Lemma 5 and Lemma 6 are NOT true.

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Algebraic Riccati Equation

The algebraic Riccati equation (A.R.E.) is

AT X + XA + XRX Q = 0 (7.20)where A, R, Q Rnn are given matrices, with R = RT , and Q = QT , andsolutions X C nn are sought. We will primarily be interested in solutionsX of (7.20), which satisfy

A + RX is Hurwitz

and are called stabilizing solutions of the Riccati equation.

Associated with the data A, R , and Q, we dene the Hamiltonian matrix

H R2n

2n

byH :=

A R

Q AT (7.21)

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Invariant Subspace Denition

Suppose A Cnn . Then (via matrix-vector multiplication), we view A asa linear map from C n C n .

A subspace V C n is said to be A-invariant if for every vector v V , thevector Av V too.

In terms of matrices, suppose that the columns of W C nm are a basis for

V . In other words, they are linearly independent, and span the space

V , so

V = {W : C m}

Then V being A-invariant is equivalent to the existence of a matrix S C mmsuch that

AW = W S

The matrix S is the matrix representation of A restricted to V , referred tothe basis W . Clearly, since W has linearly independent columns, the matrixS is unique.

If (Q, ) are the Schur decomposition of A, then for any k with 1 k n ,the rst k columns of Q span a k-dimensional invariant subspace of A, as

seen belowA [Q1 Q2] = [Q1 Q2]

11 120 22

Clearly, AQ1 = Q111.

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Eigenvectors in Invariant subspaces

Note that since S Cmm has at least one eigenvector, say = 0 m , withassociated eigenvalue , it follows that W V is an eigenvector of A,

A (W ) = AW = W S = W = (W )

Hence, every nontrivial invariant subspace of A contains an eigen-vector of A.

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Stable Invariant Subspace Denition

If S is Hurwitz, then V is a stable, A-invariant subspace.Note that while the matrix S depends on the particular choice of the basis for

V (ie., the specic columns of W ), the eigenvalues of S are only dependenton the subspace V , not the basis choice.To see this, suppose that the columns of W C nm span the same space asthe columns of W .

Obviously, there is a matrix T Cmm such that W = W T . The linearindependence of the columns of W (and W ) imply that T is invertible. Also,

A W = AW T 1 = W ST 1 = W TST 1 S

Clearly, the eigenvalues of S and S are the same.

The linear operator A|V is the transformation from V V dened as A,simply restricted to V .Since V is A-invariant, this is well-dened. Clearly, S is the matrix repre-sentation of this operator, using the columns of W as the basis choice.

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AT X + XA + XRX Q = 0 Invariant Subspace The rst theorem gives a way of constructing solutions to (7.20) in terms of invariant subspaces of H .

Theorem 69 Let V C2n

be a n-dimensional invariant subspace of H ( H is a linear map from C2n to C2n ), and let X 1 Cnn , X 2 Cnn be two complex matrices such that

V = spanX 1X 2

If X 1 is invertible, then X := X 2X 11 is a solution to the algebraic Riccati equation, and spec (A + RX ) = spec (H |V ).Proof: Since V is H invariant, there is a matrix C nn with

A R

Q AT X 1X 2

=X 1X 2

(7.22)

which givesAX 1 + RX 2 = X 1 (7.23)

andQX

1 AT X

2 = X

2 (7.24)Pre and post multiplying equation (7.23) by X 2X 11 and X 11 respec-tively gives

X 2X 11 A + X 2X 11 R X 2X 11 = X 2X 11 (7.25)and postmultiplying 7.24 by X 11 gives

Q AT X 2X 11 = X 2X 11 (7.26)Combining (7.25) and (7.26) yields

AT (X 2X 11 ) + ( X 2X 11 )A + ( X 2X 11 )R (X 2X 11 ) Q = 0so X 2X 11 is indeed a solution. Equation (7.23) implies that A+ R X 2X 11 =X 1X 11 , and hence they have the same eigenvalues. But, by denition, is a matrix representation of the map H |V , so the theorem is proved.

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AT X + XA + XRX Q = 0 Invariant Subspace The next theorem shows that the specic choice of matrices X 1 and X 2 whichspan V is not important.

Theorem 70 Let V , X 1, and X 2 be as above, and suppose there are two other matrices X 1, X 2 Cnn such that the columns of

X 1X 2

also span

V . Then X 1 is invertible if and only if X 1 is invertible. Consequently, bothX 2X 11 and X 2 X 11 are solutions to (7.20), and in fact they are equal.

Remark: Hence, the question of whether or not X 1 is invertible is a property

of the subspace, and not of the particular basis choice for the subspace.Therefore, we say that an n -dimensional subspace V has the invertibility property if in any basis, the top portion is invertible. In the literature,this is often called the complimentary property .

Proof: Since both sets of columns span the same n-dimensional subspace,there is an invertible K C nn such that

X 1X 2 =

X 1X 2 K

Obviously, X 1 is invertible if and only if X 1 is invertible, and X 2X 11 = X 2K X 1K

1= X 2 X 11 . .

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AT X + XA + XRX Q = 0 Invariant Subspace As we would hope, the converse of theorem 69 is true.

Theorem 71 If X Cn

n

is a solution to the A.R.E., then there existmatrices X 1, X 2 C nn , with X 1 invertible, such that X = X 2X 11 and the columns of

X 1X 2

span an n-dimensional invariant subspace of H .

Proof: Dene := A + RX Cnn . Multiplying this by X gives X =XA + XRX = Q AT X , the second equality coming from the fact thatX is a solution to the Riccati equation. We write these two relations as

A R

Q AT I nX

=I nX

Hence, the columns of I nX

span an n-dimensional invariant subspace

of H , and dening X 1 := I n , and X 2 := X completes the proof. .

Hence, there is a one-to-one correspondence between n-dimensional invari-ants subspaces of H with the invertibility property, and solutions of theRiccati equation.

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AT X + XA + XRX Q = 0 Eigenvalue distribution of H In the previous section, we showed that any solution X of the Riccati equa-tion is intimately related to an invariant subspace, V , of H , and the matrixA + RX is similiar (ie. related by a similarity transformation) to the re-

striction of H on V , H |V . Hence, the eigenvalues of A + RX will alwaysbe a subset of the eigenvalues of H . Recall the particular structure of theHamiltonian matrix H ,

H :=A R

Q AT Since H is real, we know that the eigenvalues of H are symmetric about thereal axis, including multiplicities. The added structure of H gives additionalsymmetry in the eigenvalues as follows.

Lemma 72 The eigenvalues of H are symmetric about the origin (and hence also about the imaginary axis).

Proof: Dene J R 2n2n by

J =0 I I 0

Note that JHJ 1 = H T . Also, JH = H T J = ( JH )T . Let p ( ) bethe characteristic polynomial of H . Then

p( ) := det ( I H )= det I JHJ 1= det I + H T

= det ( I + H )= ( 1)2ndet(I H )= p( )

(7.27)

Hence the roots of p are symmetric about the origin, and the symmetryabout the imaginary axis follows.

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AT X + XA + XRX Q = 0 Hermitian Solutions

Lemma 73 If H has no eigenvalues on the imaginary axis, then H has neigenvalues in the open-right-half plane, and n eigenvalues in the open-left-

half plane.

Theorem 74 Suppose V is a n-dimensional invariant subspace of H , and X 1, X 2 Cnn form a matrix whose colums span V . Let Cnn be a matrix representation of H

|V . If i + j

= 0 for all eigenvalues i , j

spec (), then X 1 X 2 = X 2 X 1.Proof: Using the matrix J from the proof of Lemma 72 we have that

[X 1 X 2 ] JH X 1X 2

is Hermitian, and equals ( X 1 X 2 + X 2 X 1) . Dene W := X 1 X 2 +X

2X

1. Then W

= W

, and W

+ W

= 0. By the eigenvalueassumption on , the linear map L :C nn C nn dened byL (X ) = X + X

is invertible, so W = 0.

Corollary 75 Under the same assumptions as in theorem 74, if X 1 is in-vertible, then X 2X 11 is Hermitian.

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AT X + XA + XRX Q = 0 Real Solutions Real (rather than complex) solutions arise when a symmetry condition isimposed on the invariant subspace.

Theorem 76 Let V be a n-dimensional invariant subspace of H with the invertibility property, and let X 1, X 2 Cnn form a 2n n matrix whose columns span V . Then V is conjugate symmetric ( {v : v V} = V ) if and only if X 2X 11 R nn .

Proof: Dene X := X 2X 11 . By assumption, X R nn , and

spanI

nX = V

so V is conjugate symmetric. Since V is conjugate symmetric, there is an invertible matrix K

C nn such that

X 1X 2

=X 1X 2

K

Therefore, ( X 2X 11 ) = X 2X 11 = X 2K (X 1K )1 = X 2X 11 as de-sired.

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AT X + XA + XRX Q = 0 Stabilizing Solutions Recall that for any square matrix M , there is a largest stable invariantsubspace. That is, there is a stable invariant subspace V S that contains everyother stable invariant subspace of M . This largest stable invariant subspace

is simply the span of all of the eigenvectors and generalized eigenvectorsassociated with the open-left-half plane eigenvalues. If the Schur form hasthe eigenvalues (on diagonal of ) ordered, then it is spanned by the rstset of columns of Q

Theorem 77 There is at most one solution X to the Riccati equation suchthat A + RX is Hurwitz, and if it does exist, then X is real, and symmetric.

Proof: If X is such a stabilizing solution, then by theorem 71 it can beconstructed from some n-dimensional invariant subspace of H , V . If A + RX is stable, then we must also have spec (H |V ). Recall that thespectrum of H is symmetric about the imaginary axis, so H has atmost n eigenvalues in C, and the only possible choice for V is thestable eigenspace of H . By theorem 70, every solution (if there are any)constructed from this subspace will be the same, hence there is at most1 stabilizing solution.

Associated with the stable eigenspace, any as in equation (7.22) willbe stable. Hence, if the stable eigenspace has the invertibility property,from corollary 75 we have X := X 2X 11 is Hermitian. Finally, since H is real, the stable eigenspace of H is indeed conjugate symmetric, so theassociated solution X is in fact real, and symmetric.

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AT X + XA + XRX Q = 0 domS RicIf H is (7.21) has no imaginary axis eigenvalues, then H has a unique n-dimensional stable invariant subspace V S C2n , and since H is real, thereexist matrices X 1, X 2 R nn such that

spanX 1X 2

= V S .

More concretely, there exists a matrix R nn such that

H X 1X 2

=X 1X 2

, spec () C

This leads to the following denition

Denition 78 (Denition of domS Ric) Consider H R2n2n as dened in (7.21). If H has no imaginary axis eigenvalues, and the matrices X 1, X 2R nn for which

spanX 1

X 2

is the stable, n-dimensional invariant subspace of H satisfy det(X 1) = 0 ,then H domS Ric. For H domS Ric, dene RicS (H ) := X 2X 11 , which is the unique matrix X R nn such that A + RX is stable, and

AT X + XA + XRX Q = 0 .Moreover, X = X T .

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AT X + XA + XRX Q = 0 Useful Lemma

Theorem 79 Suppose H (as in (7.21)) does not have any imaginary axis eigenvalues, R is either positive semidenite, or negative semidenite, and

(A, R ) is a stabilizable pair. Then H domS Ric.

Proof: LetX 1X 2

form a basis for the stable, n-dimensional invariant sub-

space of H (this exists, by virtue of the eigenvalue assumption aboutH ). Then

A R

Q

AT

X 1X 2

=X 1X 2

(7.28)

where Cnn is a stable matrix. Since is stable, it must be thatX 1 X 2 = X 2 X 1. We simply need to show X 1 is invertible. First we provethat Ker X 1 is -invariant, that is, the implication

x KerX 1 x KerX 1Let x KerX 1. The top row of (7.28) gives RX 2 = X 1 AX 1.Premultiply this by x X 2 to get

x X 2 RX 2x = x X 2 (X 1 AX 1) x= x X 2 X 1x= x X 1 X 2x since X 2 X 1 = X 1 X 2= 0

.

Since R 0 or R 0, it follows that RX 2x = 0. Now, using (7.28)again gives X 1x = 0, as desired. Hence KerX 1 is a subspace that is

-invariant. Now, if KerX 1 = {0}, then there is a vector v = 0 and C , Re ( ) < 0 such thatX 1v = 0

v = vRX 2v = 0

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We also have QX 1 AT X 2 = X 2, from the second row of (7.28).Mutiplying by v gives

v X 2 A I R = 0Since Re < 0, and (A, R ) is assumed stabilizable, it must be thatX 2v = 0. But X 1v = 0, and

X 1X 2

is full-column rank. Hence, v =

0, and KerX 1 is indeed the trivial set {0}. This implies that X 1 isinvertible, and hence H domS Ric. .

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AT X + XA + XRX Q = 0 Quadratic Regulator The next theorem is a main result of LQR theory.

Theorem 80 Let A,B,C be given, with (A, B ) stabilizable, and (A, C )

detectable. Dene H :=

A BB T C T C AT

Then H domS Ric. Moreover, 0 X := Ric S (H ), and

KerX

C

CA...

CA n1

Proof: Use stabilizablity of (A, B ) and detectability of ( A, C ) to show thatH has no imaginary axis eigenvalues. Specically, suppose v1, v2 Cnand R such that

A BB T

C T C

AT

v1v2

= jv1v2

Premultiply top and bottom by v2 and v1, combine, and conclude v1 =v2 = 0. Show that if ( A, B ) is stabilizable, then A, BB T is also sta-bilizable. At this point , apply Theorem 79 to conclude H domS Ric.Then, rearrange the Riccati equation into a Lyapunov equation anduse Theorem 64 to show that X 0. Finally, show that Ker X KerC ,and that Ker X is A-invariant. That shows that Ker X KerCA j forany j 0. .

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Derivatives Linear Maps

Recall that the derivative of f at x is dened to be the number lx such that

lim 0

f (x + ) f (x) lx

= 0

Suppose f : R n Rm is differentiable. Then f (x) is the unique matrixLx R mn which satises

lim 0n

1

(f (x + ) f (x) Lx ) = 0Hence, while f : Rn Rm , the value of f at a specic location x, f (x),is a linear operator from R n

R m . Hence the function f is a map from

R n L (R n , R m).Similarly, if f : H nn H nn is differentiable, then the derivative of f at xis the unique linear operator LX : H nn H nn such that satisfying

lim 0n n

1

(f (X + ) f (X ) LX ) = 0

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Derivative of Riccati Lyapunov

f (X ) := XDX A X XA C

Note,

f (X + ) = ( X + ) D (X + ) A (X + ) (X + ) A C = XDX A X XA C + DX + XD A A + D = f (X ) ( A DX ) (A DX ) + D

For any W Cnn , let LW : Hnn Hnn be the Lyapunov operator

LW (X ) := W X + XW . Using this notation,f (X + ) f (X ) LA+ DX () = D

Therefore

lim 0n n

1

[f (X + ) f (X ) LA+ DX ()] = lim 0n n1

[ D ] = 0

This means that the Lyaponov operator LA+ DX is the derivative of thematrix-valued functionXDX A X XA C

at X .

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Iterative Algorithms Roots

Suppose f : R R is differentiable. An iterative algorithm to nd a rootof f (x) = 0 is

xk+1 = xk f (xk)f (xk)

Suppose f : R n R n is differentiable. Then to rst orderf (x + ) = f (x) + Lx

An iterative algorithm to nd a root of f (x) = 0 n is obtained by settingthe rst-order approximation expression to zero, and solving for , yieldingthe update law (with xk+1 = xk + )

xk+1 = xk L1xk [f (xk)]Implicitly, it is written as

Lxk [xk+1 xk] = f (xk) (7.29)

For the Riccati equation

f (X ) := XDX

A X

XA

C

we have derived that the derivative of f at X involves the Lyapunov operator.The iteration in (7.29) appears as

LA+ DX k (X k+1 X k) = X kDX k + A X k + X kA + C In detail, this is

(A DX k) (X k+1 X k)(X k+1 X k) (A DX k) = X kDX k+ A X k+ X kA+ C which simplies (check) to

(A DX k) X k+1 + X k+1 (A DX k) = X kDX k C

Of course, questions about the well-posedness and convergence of the itera-tion must be addressed in detail.

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Identities

For any Hermitian matrices Y and Y , it is trivial to verify

Y (A DY ) + ( A DY ) Y + Y DY = Y (A D Y ) + ( A D Y ) Y + Y DY Y Y D Y Y

(7.31)

Let X denote any Hermitian solution (if one exists) to the ARE,

XDX A X XA C = 0 (7.32)This (the existence of a hermitian solution) is the departure point for allthat will eventually follow.

For any hermitian matrix Z

C = XDX + A X + XA= X (A DX ) + ( A DX ) X + XDX = X (A DZ ) + ( A DZ ) X + ZDZ (X Z ) D (X Z )

(7.33)This exploits that X solves the original Riccati equation, and uses the iden-tity (7.31), with Y = X, Y = Z .

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Iteration

Take any C C (which could be C , for instance). Also, let X 0 = X 0 0be chosen so that A DX 0 is Hurwitz. Check: Why is this possible?Iterate (if possible, which we will show is...) as

X +1 (A DX ) + ( A DX ) X +1 = X DX C (7.34)Note that by the Hurwitz assumption on A DX 0, at least the rst iterationis possible ( = 0). Also, recall that the iteration is analogous to a rst-orderalgorithm for root nding, xn+1 := xn f (xn )f (xn ) .Use Z := X in equation (7.33), and subtract from the iteration equation

(7.34), leaving(X +1 X ) (A DX ) + ( A DX ) (X +1 X )

= (X X ) D (X X ) C C (7.35)

which is valid whenever the iteration equation is valid.

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Main Induction

Now for the induction: assume X = X and A DX is Hurwitz. Notethat this is true for = 0.

Since A DX is Hurwitz, there is a unique solution X +1 to equation (7.34).Moreover, since ADX is Hurwitz, and the right-hand-side of (7.35) is 0,it follows that X +1 X 0 (note that X 0 X is not necessarily positivesemidenite).

Rewrite (7.31) with Y = X +1 and Y = X .

X +1 (A DX +1 ) + ( A DX +1 ) X +1 + X +1 DX +1= X +1 (A

DX ) + ( A

DX ) X +1 + X DX

C (X +1

X ) D (X +1

X )

Rewriting,

C = X +1 (A DX +1 ) + ( A DX +1 ) X +1+ X +1 DX +1 + ( X +1 X ) D (X +1 X )

Writing (7.33) with Z = X +1 gives

C = X (A DX +1 ) + ( A DX +1 ) X + X +1 DX +1 (X X +1 ) D (X X +1 )Subtract these, yielding

C C = ( X +1 X ) (A DX +1 ) + ( A DX +1 ) (X +1 X )+ ( X +1 X ) D (X +1 X ) + ( X X +1 ) D (X X +1 )

(7.36)

Now suppose that Re ( ) 0, and v Cn such that ( A DX +1 ) v = v .Pre and post-multiply (7.36) by v and v respectively. This gives

v C C v 0

= 2Re( )

0v (X +1 X ) v 0

+D 1/ 2 (X +1 X )D 1/ 2 (X X +1 )

v2

0Hence, both sides are in fact equal to zero, and therefore

v (X +1 X ) D (X +1 X ) = 0 .158


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Since D 0, it follows that D (X +1 X ) v = 0, which means DX +1 v =DX v. Hence

v = ( A DX +1 ) v = ( A DX ) vBut A DX is Hurwitz, and Re( ) 0, so v = 0n as desired. This meansA DX +1 is Hurwitz.Summarizing:

Suppose there exists a hermitian X = X solvingXDX A X XA C = 0

Let X 0 = X 0 0 be chosen so A DX 0 is Hurwitz Let C be a Hermitian matrix with C C . Iteratively dene (if possible) a sequence {X } =1 via

X +1 (A DX ) + ( A DX ) X +1 = X DX C The following is true:

Given the choice of X 0 and C , the sequence

{X

} =1

is well-dened,unique, and has X = X .

For each 0, A DX is Hurwitz For each 1, X X .

It remains to show that for 1, X +1 X (although it is not necessarilytrue that X 1 X 0).

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Decreasing Proof

For 1, apply (7.31) with Y = X and Y = X 1, and substitute theiteration (shifted back one step) equation (7.34) giving

X (A DX ) + ( A DX ) X + X DX = C (X X 1) D (X X 1)The iteration equation is

X +1 (A DX ) + ( A DX ) X +1 = X DX C Subtracting leaves

(X

X +1 ) (A

DX ) + ( A

DX ) (X

X +1 )

= ( X X 1) D (X X 1)The fact that A DX is Hurwitz and D 0 implies that X X +1 0.Hence, we have shown that for all 1,

X X +1 X

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Monotonic Matrix Sequences Convergence

Every nonincreasing sequence of real numbers which is bounded below hasa limit. Specically, if {xk}k=0 is a sequence of real numbers, and there is areal number such that xk

and xk+1

xk for all k, then there is a real

number x such thatlim

kxk = x

A similar result holds for symmetric matrices: Specically, if {X k}k=0 is asequence of real, symmetric matrices, and there is a real symmetric matrix such that X k and X k+1 X k for all k, then there is a real symmetricmatrix X such that

limk

X k = X

This is proven by rst considering the sequence eT i X kei k=0 , which showsthat all of the diagonal entries of the sequence have limits. Then look at(ei + e j )T X k(ei + e j ) k=0 to conclude convergence of off-diagonal terms.

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Limit of Sequence Maximal Solution

The sequence {X } =0 , generated by (7.34), has a limitlim

X =: X +

Remember that C was any hermitian matrix C .

Facts:

Since each A DX is Hurwitz, in passing to the limit we may geteigenvalues with zero real-parts, hence

maxi

Re i (A DX + ) 0

Since each X = X , it follows that X + = X + . Since each X X (where X is any hermitian solution to the equality

XDX A X XA C = 0, equation (7.32)), it follows by passing tothe limit that X + X .

The iteration equation isX +1 (A

DX ) + ( A

DX ) X +1 =

X DX

C

Taking limits yields

X + A D X + + A D X + X + = X + D X + C which has a few cancellations, leaving

X + A + A X + X + D X + + C = 0

Finally, if X

+ denotes the above limit when C

= C

, it follows that theabove ideas hold, so X + X , for all solutions X . And really nally, since X + is a hermitian solution to (7.32) with C , the

properties of X + imply that X + X + .

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Analogies with Scalar Case dx2 2ax c = 0

Matrix Statement Scalar SpecializationD = D

0 d R , with d

0

(A, D ) stabilizable a < 0 or d > 0existence of Hermitian solution a2 + cd 0Maximal solution X + x+ = a+

a 2 + cdd for d > 0

Maximal solution X + x+ = c2a for d = 0A DX + a2 + cd for d > 0A DX + a for d = 0A

DX 0 Hurwitz x0 > ad for d > 0

A DX 0 Hurwitz a < 0 for d = 0A DX Hurwitz x > ad for d > 0A DX Hurwitz a < 0 for d = 0

If d = 0, then the graph of 2ax c (with a < 0) is a straight line with apositive slope. There is one (and only one) real solution at x = c2a .If d > 0, then the graph of dx2

2ax

c is a upward directed parabola.

The minimum occurs at ad . There are either 0, 1, or 2 real solutions todx 2 2ax c = 0.

If a2 + cd < 0, then there are no solutions (the minimum is positive) If a2 + cd = 0, then there is one solution, at x = ad , which is where the

minimum occurs.

If a2 + cd > 0, then there are two solutions, at x

= a

d a 2 + cd

d and

x+ = ad + a2 + cdd which are equidistant from the minimum

ad .

LyapunovRiccati

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