Spectrum control in linear systems with incomplete feedback

ISSN 0012-2661, Differential Equations, 2009, Vol. 45, No. 9, pp. 1348–1357. c© Pleiades Publishing, Ltd., 2009.Original Russian Text c© V.A. Zaitsev, 2009, published in Differentsial’nye Uravneniya, 2009, Vol. 45, No. 9, pp. 1320–1328.

CONTROL THEORY

Spectrum Control in Linear Systemswith Incomplete Feedback

V. A. ZaitsevUdmurt State University, Izhevsk, Russia

Received October 1, 2007

Abstract— For a linear stationary control system closed by a linear incomplete feedback,we obtain necessary and sufficient conditions for the solvability of the spectrum control problemin the case of coefficients of special form.

DOI: 10.1134/S0012266109090109

1. NOTATION AND DEFINITIONS

Let e1, . . . , en be the canonical basis in the space Rn [i.e., e1 = col(1, 0, . . . , 0), . . . , en =

col(0, . . . , 0, 1)]; let Mm,n be the space of real m×n matrices; let Mn := Mn,n; let I = [e1, . . . , en] ∈Mn be the identity matrix; let ∗ be the operation of transposition of a vector or a matrix; letJ0 := I; let J1 be the matrix whose first superdiagonal is all ones and whose other entries are zero(i.e., J1 :=

∑n−1

i=1 eie∗i+1 ∈ Mn); let Jk := Jk

1 , k ∈ N (thus, Jk = 0 ∈ Mn for k ≥ n); let χ(A;λ) bethe characteristic polynomial of a matrix A; and let SpA be the trace of a matrix A.

Consider the linear stationary control system

x = Ax + Bu, (x, u) ∈ Rn × R

m, (1)y = C∗x, y ∈ R

k, (2)

specified by a matrix (A,B,C) ∈ Mn,n+m+k. Let the control in system (1), (2) be constructed bythe principle of linear incomplete feedback in the form u = Uy, where U ∈ Mm,k is a constantmatrix. The corresponding closed system has the form

x = (A + BUC∗)x, x ∈ Rn. (3)

In the present paper, we consider the spectrum control problem for the matrix A + BUC∗ ofsystem (3). This problem can be stated in different ways.

(i) There are given matrices A, B, and C (possibly, complex). For an arbitrary set {μj}nj=1

of complex numbers, construct a matrix (possibly, complex) control U such that the eigenvaluesλj(A + BUC∗) of the matrix A + BUC∗ coincide with the numbers μj.

(ii) There are given real matrices A, B, and C. For an arbitrary set {μj}nj=1 of complex numbers

closed with respect to complex conjugation, construct a real matrix U such that λj(A+BUC∗) = μj,j = 1, . . . , n.

We assume that A, B, and C are real matrices and solve this problem in the second setting; allobtained assertions and proofs hold for the first setting as well.

Definition 1. We say that the spectrum control problem for the matrix A + BUC∗ is solvableif, for any polynomial p(λ) = λn +γ1λ

n−1 + · · ·+γn of degree n with real coefficients γi, there existsa real matrix U ∈ Mm,k such that the characteristic polynomial χ(A + BUC∗;λ) of the matrixA + BUC∗ coincides with p(λ).

This problem is also referred to as the eigenvalue placement problem [1, p. 159] or the modalcontrol problem [2, p. 435]. If C = I, then the spectrum control problem is solvable if and only

1348

SPECTRUM CONTROL IN LINEAR SYSTEMS WITH INCOMPLETE FEEDBACK 1349

if system (1) is completely controllable [3, p. 320; 4]. This problem with m < n and k < n wasstudied by numerous authors. A detailed review of known necessary and sufficient conditions forthe solvability of this problem can be found in [1, Chap. III, Sec. 7]. In the present paper, we obtainnew conditions for the solvability of this problem for system (3) with coefficients of the form (4)(see below), which are necessary and sufficient conditions. These results generalize those in [5, 6].

2. RESULTS

Let the coefficients of system (3) have the following form:

A = {aij}ni,j=1, ai,i+1 �= 0, i = 1, . . . , n − 1; aij = 0, j > i + 1;

B = {bij}, C = {cil}, i = 1, . . . , n, j = 1, . . . ,m, l = 1, . . . , k;bij = 0, i = 1, . . . , p − 1, j = 1, . . . ,m;cil = 0, i = p + 1, . . . , n, l = 1, . . . , k; p ∈ {1, . . . , n}.

(4)

Let λn +α1λn−1 + · · ·+αn := χ(A;λ). Set α0 := 1. We remove the last row from the matrix A and

denote the resulting matrix by Q ∈ Mn−1,n. Let us construct the matrix S1 =∥∥∥∥

e∗1Q

∥∥∥∥. Then S1 ∈ Mn

is a lower triangular matrix, and detS1 �= 0. For each l = 2, . . . , n − 1, on the basis of the matrixSl−1 = {sl−1

ij }ni,j=1, we construct the matrix Sl = {sl

ij}ni,j=1 as follows: sl

11 := 1, sl1j := sl

j1 := 0,j = 2, . . . , n; sl

ij = sl−1i−1,j−1, i, j = 2, . . . , n. Then the Sl are lower triangular nonsingular matrices

for all l = 1, . . . , n − 1. Let S = Sn−1 · · ·S1. Then S is a lower triangular nonsingular matrix aswell. Let us construct the matrix

G :=n∑

i=1

αi−1J∗i−1.

Theorem 1. Let system (3) with matrices of the form (4) be given, and let

λn + γ1λn−1 + · · · + γn := χ(A + BUC∗;λ).

Then the coefficients γi of the characteristic polynomial of the matrix A + BUC∗ can be expressedvia the coefficients of system (3), (4) as follows :

γi = αi − Sp(SBUC∗S−1Ji−1G), i = 1, . . . , n. (5)

Theorem 2. The spectrum control problem for system (3) with matrices of the form (4) issolvable if and only if the matrices

C∗S−1J0GSB, C∗S−1J1GSB, . . . , C∗S−1Jn−1GSB (6)

are linearly independent.

Proof of Theorem 2. We prove Theorem 2 as a corollary of Theorem 1 and explicitly writeout a control reducing the characteristic polynomial to a given form. Let

p(λ) = λn + γ1λn−1 + · · · + γn

be a given polynomial with real coefficients γi. The problem is to construct U such thatχ(A + BUC∗;λ) = p(λ), i.e., such that relations (5) hold:

γi = αi − Sp(SBUC∗S−1Ji−1G) = αi − Sp(UC∗S−1Ji−1GSB), i = 1, . . . , n.

It is a system of n linear equations for the mk unknown entries upq, p = 1, . . . ,m, q = 1, . . . , k,of the matrix U . We introduce the mapping vec : Mk,m → R

km, which “expands” the matrixH = {hij}, i = 1, . . . , k, j = 1, . . . ,m, by rows into the column vector

vec H := col(h11, . . . , h1m, . . . , hk1, . . . , hkm).

DIFFERENTIAL EQUATIONS Vol. 45 No. 9 2009

1350 ZAITSEV

Obviously, for any matrices H1,H2 ∈ Mk,m, one has Sp(H∗1H2) = (vec H1)∗(vec H2). Then the

system of linear equations (5) can be rewritten in the vector form

α − P ∗v = γ. (7)

Here

P = [vec C∗S−1J0GSB, . . . , vec C∗S−1Jn−1GSB] ∈ Mkm,n, α = col(α1, . . . , αn) ∈ Rn,

γ = col(γ1, . . . , γn) ∈ Rn, v = vec U∗ ∈ R

km.

If the matrices (6) are linearly independent, then rankP = n. Then P ∗P is nonsingular, sothat system (7) is solvable for any γ and, in particular, has the solution v = P (P ∗P )−1(α − γ).Therefore, the spectrum control problem is solvable, and the corresponding control has the formU = (vec −1v)∗. If the matrices (6) are linearly dependent, then rankP < n and system (7) isunsolvable whenever γ = α − β, where β /∈ ImP ∗; consequently, the spectrum control problemis unsolvable. The proof of the theorem is complete.

Remark 1. It follows from Theorem 2 that a necessary condition for the solvability of thespectrum control problem for system (3) with matrices (4) is that mk ≥ n. If the matrices (6)are linearly independent and mk = n, then the control U ensuring that χ(A + BUC∗;λ) = p(λ) isunique; if the matrices (6) are linearly independent and mk > n, then such a control is nonunique.

If only l < n of the matrices (6) are linearly independent, then the spectrum control problemis unsolvable. Nevertheless, the spectrum can be partly controlled; namely, one can ensure thatχ(A + BUC∗;λ) = p(λ) for any γ = (γ1, . . . , γn) in the l-dimensional linear manifold α − Im P ∗

in Rn.

Corollary 1. If the matrices (6) are linearly independent , then system (3), (4) is stabilizable inthe class of constant matrix controls U ; i.e., for any κ > 0, there exists a constant matrix U suchthat the eigenvalues λi of the matrix A+BUC∗ satisfy the conditions Reλi ≤ −κ < 0, i = 1, . . . , n.

Remark 2. Theorems 1 and 2 hold for system (3) with coefficients of the form (4). Suchsystems do not exhaust all possibly systems of the form (3). Nevertheless, the class of such systemsis sufficiently large; in particular, it contains the following important class of systems. Considera plant described by a linear ordinary differential equation of order n with constant coefficients inwhich a linear combination of m signals and their derivatives of order ≤ n − p is supplied to theinput and k distinct linear combinations of the plant state z and its derivatives of order ≤ p−1 aremeasured (see [6]). Let us construct a linear incomplete feedback control. By using the standardsubstitution z = x1, z = x2, . . . , z(n−1) = xn, we pass from an equation of order n to a system ofdifferential equations. Then the coefficients of the resulting system of equations have exactly theform (4). The corresponding result, Theorem 2 for an ordinary differential equation, was obtainedin [6].

3. PROOF OF THEOREM 1

Let Q′ ∈ Mn−1 be the matrix obtained from S1 by removing the last row and the last column.We introduce the matrix A1 := S1AS−1

1 .

Lemma 1. The matrix A1 has the form A1 =∥∥∥∥

0 | Q′

∗ ∗ ∗

∥∥∥∥; here 0 ∈ R

n−1, and the last row

contains numbers such that χ(A;λ) = χ(A1;λ).

In this lemma, we claim the following. If we multiply the matrix A by S1 on the left and by S−11

on the right, then the rectangular “support block” Q ∈ Mn−1,n of the matrix A is shifted right anddown along the diagonal; the last row and the last column of the matrix Q are lost, and the firstcolumn (of height n−1) consisting of zeros and the first row e∗2 (of length n) are added; the last rowof the matrix A changes so as to ensure that the characteristic polynomial is preserved. Lemma 1 iswell stated, since the last row of the matrix A1 can be uniquely reconstructed from its characteristicpolynomial. This is a consequence of the following auxiliary assertion.



Lemma 2. Let P =∥∥∥∥

Q

ξ

∥∥∥∥ and R =

∥∥∥∥

Q

ψ

∥∥∥∥ be matrices such that the characteristic polynomials

of these matrices coincide, χ(P ;λ) = χ(R;λ). Then the last rows of these matrices coincide.

Proof. Let ξ = (p1, . . . , pn) ∈ Rn∗ and ψ = (r1, . . . , rn) ∈ R

n∗. Consider the matrix λI − P .By Δi, i = 1, . . . , n, we denote the principal diagonal minors of this matrix, Δ1 = λ − a11, Δ2 =(λ− a11)(λ− a22)− a12a21, . . . , Δn = det(λI −P ). Note that, for all i = 1, . . . , n, the degree of thepolynomial Δi is equal exactly to i, and the leading coefficient of λi is equal to unity. We expanddet(λI − P ) with respect to the last row; then we obtain

χ(P ;λ) = (λ − pn)Δn−1 − (−pn−1)(−an−1,n)Δn−2 + (−pn−2)(−an−1,n)(−an−2,n−1)Δn−3 + · · ·+ (−1)n−2(−p2)(−an−1,n) · · · (−a23)Δ1 + (−1)n−1(−p1)(−an−1,n) · · · (−a12)

= (λ − pn)Δn−1 − pn−1an−1,nΔn−2 − pn−2an−1,nan−2,n−1Δn−3 − · · ·− p2an−1,n · · · a23Δ1 − p1an−1,n · · · a12.

Likewise,

χ(R;λ) = (λ − rn)Δn−1 − rn−1an−1,nΔn−2 − rn−2an−1,nan−2,n−1Δn−3 − · · ·− r2an−1,n · · · a23Δ1 − r1an−1,n · · · a12.

The characteristic polynomials of the matrices coincide. Let us subtract the first polynomial fromthe second one; then we obtain

(pn − rn)Δn−1 + (pn−1 − rn−1)an−1,nΔn−2 + · · ·+ (p2 − r2)an−1,n · · · a23Δ1 + (p1 − r1)an−1,n · · · a12 = 0.

The left-hand side is a polynomial of degree ≤ n − 1 in λ, the right-hand side is zero, and con-sequently, all coefficients of λi−1, i = 1, . . . , n, are zero. Since the coefficient of λn−1 is zero, andof the polynomials Δ1, . . . ,Δn−1 only the polynomial Δn−1 has degree n − 1, i.e., contains themonomial λn−1, it follows that the coefficient of Δn−1 is zero; consequently, pn = rn. We proceedin a similar way. Since ak,k+1 �= 0 and all polynomials Δi have distinct degrees, we have pi = ri forall i = 1, . . . , n. The proof of the lemma is complete.

Proof of Lemma 1. Since S1 =∥∥∥∥

e∗1

Q

∥∥∥∥, we have QS−1

1 = ‖ 0 | I ‖ ∈ Mn−1,n, 0 ∈ Rn−1,

I ∈ Mn−1. Therefore, AS−11 =

∥∥∥∥

Q

∗

∥∥∥∥ · S−1

1 =∥∥∥∥

0 | I

∗ ∗ ∗

∥∥∥∥. Since S1 =

∥∥∥∥

Q′ | 0∗ ∗ ∗

∥∥∥∥, it follows

that

S1AS−11 =

∥∥∥∥∥

Q′ | 0∗ ∗ ∗

∥∥∥∥∥·∥∥∥∥∥

0 | I

∗ ∗ ∗

∥∥∥∥∥

=

∥∥∥∥∥

0 | Q′

∗ ∗ ∗

∥∥∥∥∥

.

The proof of the lemma is complete.The matrix A1 in Lemma 1 has the form of the matrix A in (4). (All entries lying above

the superdiagonal are zero, and the superdiagonal consists of nonzero entries.) In the matrix A1,we remove the last row and add the first row e∗1. Then we obtain the matrix S2. We constructthe matrix A2 = S2A1S

−12 = S2S1AS−1

1 S−12 and use Lemma 1. As a result, the “support block”

is again shifted right and down along the diagonal, the characteristic polynomial is preserved,

the matrix A2 has the form A2 =∥∥∥∥

0 | Q′1

∗ ∗ ∗

∥∥∥∥; here 0 ∈ R

n−1, and the matrix Q′1 ∈ Mn−1 is

obtained from the matrix S2 by removing the last row and the last column. By using Lemma 1

n − 1 times, we obtain the matrix An−1 = Sn−1 · · ·S1AS−11 · · ·S−1

n−1 =∥∥∥∥

0 | I

∗ ∗ ∗

∥∥∥∥, 0 ∈ R

n−1,


1352 ZAITSEV

I ∈ Mn−1; moreover, χ(An−1;λ) = χ(A;λ). Consequently, the last row of the matrix An−1 containsthe coefficients αi of the characteristic polynomial of the matrix A, and the matrix An−1 is thecompanion matrix for the polynomial χ(A;λ). We set A = SAS−1, where S = Sn−1 · · ·S1, andϕ = (−αn,−αn−1, . . . ,−α1) ∈ R

n∗. Then the following assertion holds.

Lemma 3. The matrix A has the form

A = J1 + enϕ. (8)

Now let us construct the matrices B := SB and C∗ = C∗S−1. We have

χ(A + BUC∗;λ) = χ(S(A + BUC∗)S−1;λ) = χ(A + BUC∗;λ). (9)

Further, we use the following lemma, whose proof will be given below.

Lemma 4. Let A have the form (8), let D ∈ Mn, and let

D =

∥∥∥∥∥

0 | 0F | 0

∥∥∥∥∥

, F ∈ Mn−p+1,p; (10)

here p ∈ {1, . . . , n}. Let χ(A + D;λ) = λn + γ1λn−1 + · · · + γn. Then γi = αi − Sp(DJi−1G) for

all i = 1, . . . , n.

Since S is a lower triangular matrix, it follows that the matrices B and C have the same formas B and C; i.e., the first p − 1 rows of the matrix B and the last n − p rows of the matrix C arezero. This implies that the matrix BUC∗ has the form (10) of the matrix D, that is, the block form∥∥∥∥

0 0F 0

∥∥∥∥, where the right upper corner entry of the matrix F lies on the main diagonal. Then it

follows from Lemma 4 that if χ(A + BUC∗;λ) = λn + γ1λn−1 + · · · + γn, then the relations

γi = αi − Sp(BUC∗Ji−1G) = αi − Sp(SBUC∗S−1Ji−1G)

hold for all i = 1, . . . , n. By virtue of relation (9), relations (5) are true, which completes the proofof Theorem 1. It remains to prove Lemma 4.

4. PROOF OF LEMMA 4

Let us prove preliminarily an auxiliary assertion.

Lemma 5. Let

H1,H2 ∈ Mn : H1 =

∥∥∥∥∥∥∥

0 | 0H | 0ξ | 0

∥∥∥∥∥∥∥

, H2 =

∥∥∥∥∥

0 | 0 | 00 | H | 0

∥∥∥∥∥

be two given matrices, where H =

∥∥∥∥∥∥∥

hp1 . . . hpp

. . . . . . . . .

hn−1,1 . . . hn−1,p

∥∥∥∥∥∥∥

∈ Mn−p,p, ξ = (hn1, . . . , hnp) ∈ Rp∗,

p ∈ {1, . . . , n − 1}. Then

Sp(H1JsG) = Sp(H2JsG) for all s = 0, . . . , n − p − 1, (11)

Sp(H1JsG) = Sp(H2JsG) +n∑

i=p

αn−ihi,n−s for all s = n − p, . . . , n − 1; (12)



i.e.,

Sp(H1J0G) = Sp(H2J0G), . . . , Sp(H1Jn−p−1G) = Sp(H2Jn−p−1G),Sp(H1Jn−pG) = Sp(H2Jn−pG) + (α0hnp + α1hn−1,p + · · · + αn−phpp),

Sp(H1Jn−p+1G) = Sp(H2Jn−p+1G) + (α0hn,p−1 + α1hn−1,p−1 + · · · + αn−php,p−1), . . . ,

Sp(H1Jn−1G) = Sp(H2Jn−1G) + (α0hn1 + α1hn−1,1 + · · · + αn−php1).

(13)

Proof. We represent the matrix H1 as the sum H1 = H ′ + H ′′, where

H ′ =

∥∥∥∥∥∥∥

0 | 0H | 00 | 0

∥∥∥∥∥∥∥

, H ′′ =

∥∥∥∥∥∥∥

0 | 00 | 0ξ | 0

∥∥∥∥∥∥∥

.

Then Sp(H1JsG) = Sp(H ′JsG)+Sp(H ′′JsG). First, we find Sp(H ′′JsG). Note that J∗kei = ek+i for

k + i ≤ n and J∗kei = 0 ∈ R

n for k + i > n; in addition, Sp(eie∗j ) = δij , where δij is the Kronecker

delta. The matrix H ′′ can be represented in the form H ′′ =∑p

j=1 hnjene∗j . We have

Sp(H ′′JsG) = Sp(GH ′′Js) = Sp

(n−1∑

k=0

αkJ∗k

p∑

j=1

hnjene∗jJs

)

= Sp

(p∑

j=1

hnj

(n−1∑

k=0

αkJ∗ken

)

(e∗jJs)

)

= Sp

(p∑

j=1

hnjα0ene∗j+s

)

=: κ1.

If s ≤ n − p − 1, then j + s ≤ n − 1, since j ≤ p. Therefore, Sp(ene∗j+s) = 0 for any j ≤ p;consequently, κ1 = 0. If s ∈ {n − p, . . . , n − 1}, then j + s coincides with n for j = n − s, andSp(ene∗j+s) = 0 for the remaining j ∈ {1, . . . , p}\{n− s}. Therefore, κ1 = hn,n−sα0. Thus, we have

Sp(H ′′JsG) = 0 for all s = 0, . . . , n − p − 1, (14)Sp(H ′′JsG) = hn,n−sα0 for all s = n − p, . . . , n − 1. (15)

Now consider the matrices H2 and H ′. The matrix H ′ can be represented in the form

H ′ =n−1∑

i=p

p∑

j=1

hijPij ,

where Pij = eie∗j and

H2 =n−1∑

i=p

p∑

j=1

hijPi+1,j+1.

Let us clarify how different Sp(PijJsG) and Sp(Pi+1,j+1JsG) are for various values of s. We have

κ2 := Sp(PijJsG) = Sp(GPijJs) = Sp

(n−1∑

k=0

αkJ∗k eie

∗jJs

)

= Sp

(n−1∑

k=0

αkek+ie∗j+s

)

=n−1∑

k=0

αkδk+i,j+s.

If s ∈ {0, . . . , i− j − 1}, then j + s < i ≤ k + i; consequently, κ2 = 0. If s ∈ {i− j, . . . , n− j}, then0 ≤ s − i + j ≤ n − i ≤ n − 1. Then the relation k + i = j + s is true for k = s − i + j; therefore,κ2 = αs−i+j. If s ∈ {n− j +1, . . . , n−1}, then j +s > n; consequently, e∗jJs = 0, whence we obtainκ2 = 0. Further,

κ3 := Sp(Pi+1,j+1JsG) = Sp(GPi+1,j+1Js) = Sp

(n−1∑

k=0

αkJ∗kei+1e

∗j+1Js

)

=n−1∑

k=0

αkδk+i+1,j+s+1.


1354 ZAITSEV

If s ∈ {0, . . . , i − j − 1}, then j + s + 1 < i + 1 ≤ k + i + 1; consequently, κ3 = 0. If s ∈{i − j, . . . , n − j − 1}, then 0 ≤ s − i + j ≤ n − i − 1 ≤ n − 1. In this case, the relationk + i + 1 = j + s + 1 holds for k = s− i + j; therefore, κ3 = αs−i+j . If s ∈ {n − j, . . . , n − 1}, thenj + s + 1 > n; consequently, e∗j+1Js = 0, whence we obtain κ3 = 0. Thus, we have

Sp(PijJsG) = Sp(Pi+1,j+1JsG) = 0, s ∈ {0, . . . , i − j − 1},Sp(PijJsG) = Sp(Pi+1,j+1JsG) = αs−i+j , s ∈ {i − j, . . . , n − j − 1},Sp(PijJsG) = αn−i, Sp(Pi+1,j+1JsG) = 0, s = n − j,

Sp(PijJsG) = Sp(Pi+1,j+1JsG) = 0, s ∈ {n − j + 1, . . . , n − 1}.

(16)

Therefore, Sp(PijJsG) and Sp(Pi+1,j+1JsG) are different only for s = n − j.Let s ∈ {0, . . . , n − p − 1}. Then s ≤ n − p − 1 ≤ n − j − 1 for all j ∈ {1, . . . , p}. By virtue of

the first two rows in (16), we have

n−1∑

i=p

p∑

j=1

hij Sp(PijJsG) =n−1∑

i=p

p∑

j=1

hij Sp(Pi+1,j+1JsG);

consequently, Sp(H ′JsG) = Sp(H2JsG), and, by (14), we obtain (11).Now let s ∈ {n − p, . . . , n − 1}. The value s = n − j gets in this interval as j runs from 1 to p.

By (16), we have

p∑

j=1

hij Sp(PijJsG) =p∑

j=1

hij Sp(Pi+1,j+1JsG) + hi,n−sαn−i

=⇒n−1∑

i=p

p∑

j=1

hij Sp(PijJsG) =n−1∑

i=p

p∑

j=1

hij Sp(Pi+1,j+1JsG) +n−1∑

i=p

hi,n−sαn−i.

Hence it follows that Sp(H ′JsG) = Sp(H2JsG) +∑n−1

i=p hi,n−sαn−i. By adding the last relation toformula (15), we obtain (12). The proof of Lemma 5 is complete.

Let us proceed with the proof of Lemma 4. Consider the index k of the row in which the rightupper corner entry of the left lower block F of the matrix D is located. Let us make the proof byinduction on k changing from n to 1. The basis of induction is the following: let k = n. Then

D =∥∥∥∥

0η

∥∥∥∥, η = (dn1, . . . , dnn) ∈ R

n∗; consequently,

χ(A + D;λ) = λn + (α1 − dnn)λn−1 + · · · + (αn − dn1).

Hence we have γi = αi − dn,n−i+1. Further,

Sp(DJi−1G) = Sp(GDJi−1) = Sp

(n−1∑

k=0

αkJ∗k

n∑

j=1

dnjene∗jJi−1

)

= Sp

(n∑

j=1

dnj

(n−1∑

k=0

αkJ∗k en

)

(e∗jJi−1)

)

= Sp

(n∑

j=1

dnjα0ene∗j+i−1

)

=n∑

j=1

dnjδn,j+i−1 = dn,n−i+1

for all i = 1, . . . , n.



The proof of the basis is complete. The inductive assumption is the following: let the assertionof the lemma hold for any k ∈ {p + 1, . . . , n}. Let us show that it holds for k = p as well. Considerthe matrix D in (10). By the inductive assumption, p < n. We introduce the following notation:

L =

∥∥∥∥∥∥∥

dp1 . . . dpp

· · · · · · · · ·dn−1,1 . . . dn−1,p

∥∥∥∥∥∥∥

∈ Mn−p,p, σ = (dn1, . . . , dnp) ∈ Rp∗,

ψ = (dn1, . . . , dnp, 0, . . . , 0) ∈ Rn∗.

Then D =

∥∥∥∥∥∥∥

0 | 0L | 0σ | 0

∥∥∥∥∥∥∥

. We represent D in the form D = D′ + D′′, where

D′ =

∥∥∥∥∥∥∥

0 | 0L | 00 | 0

∥∥∥∥∥∥∥

, D′′ =

∥∥∥∥∥∥∥

0 | 00 | 0σ | 0

∥∥∥∥∥∥∥

.

Let D := J∗1 D =

∥∥∥∥

0 0L 0

∥∥∥∥ ∈ Mn and D := DJ1 =

∥∥∥∥

0 | 0 | 00 | L | 0

∥∥∥∥ ∈ Mn. Further, we set

K :=∥∥∥∥

0ϕ

∥∥∥∥ ∈ Mn, where ϕ = (−αn, . . . ,−α1) ∈ R

n∗. Then A = J1 + K. The matrix A + D

is a matrix of the form (4). For the matrix A + D, we construct the matrix T by analogy withthe construction of the matrix S1 for A; namely, from A + D, we remove the last row and add

the first row e∗1. Then T = I + D =∥∥∥∥

Ip 0L In−p

∥∥∥∥, Ip ∈ Mp, In−p ∈ Mn−p. Hence it follows that

T−1 = I − D. We multiply the matrix T−1 by the matrix (A + D) on the right and the matrix Ton the left. We have

(A + D)T−1 = (A + D)(I − D) = A + D − AD − DD.

The first p rows of the matrix D are zero, and the last n − p columns of the matrix D are zero;therefore, DD = 0. Further, AD = (J1 + K)D = J1D + KD, but J1D = D′; consequently,D − AD = D −D′ − KD = D′′ − KD. Therefore, (A + D)T−1 = A + D′′ − KD = A + K1, where

K1 =∥∥∥∥

0ζ

∥∥∥∥ ∈ Mn and ζ = ψ − ϕD ∈ R

n∗. Next, we have

T (A + D)T−1 = (I + D)(A + K1) = A + K1 + DA + DK1.

Since p < n, it follows that the last column of the matrix D is zero; and since the first n − 1 rowsof the matrix K1 are zero, we have DK1 = 0. Further,

DA = D(J1 + K) = DJ1 + DK.

We have DJ1 = D and DK = 0, since the first n− 1 rows of the matrix K are zero. Consequently,T (A + D)T−1 = A + K1 + D. Set A := A + K1. Then

T (A + D)T−1 = A + D.


1356 ZAITSEV

Further, A = J1 + K + K1; i.e., the matrix A has the same form as the matrix A and differsfrom it only by the last row. Let ϕ = (−αn, . . . ,−α1) ∈ R

n∗ be the last row of the matrix A.Then ϕ = ϕ + ζ = ϕ + ψ − ϕD. By multiplying both sides of the last relation by −1, we obtain−ϕ = −ϕ − ψ + ϕD. We rewrite this relation in terms of coordinates starting from the lastcoordinate. We obtain

α1 = α1, . . . , αn−p = αn−p,

αn−p+1 = αn−p+1 − dnp − α1dn−1,p − · · · − αn−pdpp = αn−p+1 −n∑

i=p

αn−idip, . . . ,

αn = αn − dn1 − α1dn−1,1 − α2dn−2,1 − · · · − αn−pdp1 = αn −n∑

i=p

αn−idi1.

(17)

Letλn + γ1λ

n−1 + · · · + γn := χ(A + D;λ).

The matrices A + D and A + D are similar; therefore,

χ(A + D;λ) = χ(A + D;λ).

By the inductive assumption, the assertion of Lemma 4 holds for the matrix A + D since the rightupper corner entry of the left lower block ‖ 0 L ‖ of the matrix D lies on the diagonal in the(p + 1)st row. Consequently,

γ1 = α1 − Sp(DJ0G), . . . , γn = αn − Sp(DJn−1G). (18)

The matrix D has the form of the matrix H1; and D has the form of the matrix H2 in Lemma 5.Therefore, by (13), we have the relations

Sp(DJ0G) = Sp(DJ0G), . . . , Sp(DJn−p−1G) = Sp(DJn−p−1G),

Sp(DJn−pG) = Sp(DJn−pG) + (α0dnp + α1dn−1,p + · · · + αn−pdpp), . . . ,

Sp(DJn−1G) = Sp(DJn−1G) + (α0dn1 + α1dn−1,1 + · · · + αn−pdp1).

(19)

By substituting (17) and (19) into (18), we obtain

γ1 = α1 − Sp(DJ0G) = α1 − Sp(DJ0G), . . . ,

γn−p = αn−p − Sp(DJn−p−1G) = αn−p − Sp(DJn−p−1G),

γn−p+1 = αn−p+1 − Sp(DJn−pG) =

(

αn−p+1 −n∑

i=p

αn−idip

)

−(

Sp(DJn−pG) −n∑

i=p

αn−idip

)

= αn−p+1 − Sp(DJn−pG), . . . ,

γn = αn − Sp(DJn−1G) =

(

αn −n∑

i=p

αn−idi1

)

−(

Sp(DJn−1G) −n∑

i=p

αn−idi1

)

= αn − Sp(DJn−1G).

The proof of Lemma 4 is complete.

ACKNOWLEDGMENTS

The research was supported by the Russian Foundation for Basic Research (projects nos. 06-01-00258 and 09-01-403).



REFERENCES

1. Leonov, G.A. and Shumafov, M.M., Metody stabilizatsii lineinykh upravlyaemykh sistem (StabilizationMethods for Linear Control Systems), St. Petersburg, 2005.

2. Spravochnik po teorii avtomaticheskogo regulirovaniya (Reference Book on Automatic Control Theory),Krasovskii, A.A., Ed., Moscow, 1987.

3. Popov, V.M., Giperustoichivost’ avtomaticheskikh sistem (Hyperstability of Automatic Systems),Moscow: Nauka, 1970.

4. Wonham, W.M., On Pole Assignment in Multi-Input Controllable Linear Systems, IEEE Trans. Auto-mat. Control , 1967, vol. AC-12, no. 6, pp. 660–665.

5. Zaitsev, V.A., On the Control of Lyapunov Characteristic Exponents of a Stationary System withan Observer, Differ. Uravn., 2000, vol. 36, no. 6, pp. 855–856.

6. Zaitsev, V.A., Modal Control of a Linear Differential Equation with Incomplete Feedback, Differ.Uravn., 2003, vol. 39, no. 1, pp. 133–135.


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