The end of the saga? Solution to a problem posed by R. Sommer · The end of the saga? Solution to a...
Transcript of The end of the saga? Solution to a problem posed by R. Sommer · The end of the saga? Solution to a...
The end of the saga? Solution to a problemposed by R. Sommer
Hannes GernandtTU Ilmenau
Siegmundsburg @ Ilmenau,
29.08.17
Perturbation of DAEs
Consider a DAEs ddtEx(t) = Ax(t) with eigenvalues close to
imaginary axis. Under 100000 random perturbations(E ,A)→ (E + xyT ,A) with ‖x‖ = ‖y‖ = 1 we obtain:
What to do? A First Method!
Assume that we have a linear time invariant system,
Ed
dtx(t) = Ax(t) + b · u(t), x(0) = x0
then we could apply a state feedback u(t) = f T x(t), f ∈ Rn
and obtain
Ed
dtx(t) = Ax(t) + b · f T x(t) = (A + bf T )x(t), x(0) = x0.
In this case the problem from the last slide can be solved bydoing applying the classical pole placement.
Second Method: Network redesign (Sommer et. al. )
Method: Add step-by-step capacitances to the network,described by d
dt cij(ei − ej)(ei − ej)T , cij > 0.
New DAE: ddt (E + cij(ei − ej)(ei − ej)
T )x(t) = Ax(t)
What have both methods in common?
We reformulate the DAE as a matrix pencilA(s) = sE − A ∈ Rn×n[s] then the feedback problem can bewritten as
sE − (A + bf T )
and the redesign problem as
s(E + cij(ei − ej)(ei − ej)T )− A.
Connection: Both are rank one perturbationssE − A + (αs − β)xyT of the matrix pencil.
Therefore we study the following problem:
Given: A matrix pencil sE − A, E ,A ∈ Rn×n.
Task: Move the eigenvalues into a certain region G ⊂ C.
First Question: Which eigenvalue sets can be obtained underrank one perturbations?
Some notations
From now on we consider the matrix pencil A(s) = sE − A
The spectrum of A is given by
σ(A) := {λ ∈ C | det(λE − A) = 0}, for E invertible and
σ(A) := {λ ∈ C | det(λE − A) = 0} ∪ {∞}, for E singular.
A(s) = sE − A regular :⇐⇒ det(sE − A) 6= 0
The algebraic multiplicity amA(λ) is the multiplicity of λ as aroot of det(sE − A)
The geometric multiplicity is given bygmA(λ) := dim ker(λE − A) for λ 6=∞ andgmA(∞) := dim ker E
1. Placement in the general case
Change of the algebraic and geometric multiplicity
Denote by m(λ) the length of the longest JC at λ ∈ σ(A) anddefine
M(A) :=∑
λ∈σ(A)m(λ)
Proposition (G, Trunk ’17)
Let sE − A be regular and P be of rank one such that A+ P isregular, then the following estimates hold:
amA(λ)−m(λ) ≤ amA+P(λ), for λ ∈ σ(A),
amA+P(λ) ≤ amA(λ) + M(A)−m(λ), for λ ∈ σ(A),
0 ≤∑
µ∈C\σ(A)
amA+P(µ) ≤ M(A),
|gmA+P(µ)− gmA(µ)| ≤ 1, for µ ∈ C ∪ {∞}.
Rank one eigenvalue placement problem
Given A(s) = sE − A withσ(A) =
..{λ1, . . . , λm}m(λ1), . . . ,m(λm)
and
pws. dist. µ1, . . . , µn ∈ CFind x , y ∈ Rn withP(s) = (αs − β)xyT suchthatσ(A+ P) = {µ1, . . . , µn}.
Eigenvalue placement for regular pencils
Theorem (G, Trunk ’17)
Given A(s) = sE − A regular and pairwise distinctµ1, . . . , µM(A) ∈ C symmetric w.r.t the real line, then we find
P(s) = (αs − β)uvT , α, β ∈ R u, v ∈ Rn
such that
σ(A+ P) = {µ1, . . . , µM(A)} ∪ {λ ∈ σ(A) | gmA(λ) ≥ 2}.and
amA+P(λ) =
amA(λ)−m(λ) + 1, for λ = µi ∈ σ(A),
amA(λ)−m(λ), for λ ∈ σ(A) \ {µi}M(A)i=1 ,
1, for λ = µi /∈ σ(A),
0, for λ /∈ σ(A) ∪ {µi}M(A)i=1 .
Consequence for matrices
For A ∈ Rn×n with gm A(λ) ≤ 1 and µ1, . . . , µn ∈ C symmetricw.r.t to the real line then there exist u, v ∈ Rn such thatσ(A + uvT ) = {µ1, . . . , µn}.
Im
Re
Open Question:How can we deter-mine u, v ∈ Rn andare they unique?
Consequence for matrices
For A ∈ Rn×n with gm A(λ) ≤ 1 and µ1, . . . , µn ∈ C symmetricw.r.t to the real line then there exist u, v ∈ Rn such thatσ(A + uvT ) = {µ1, . . . , µn}.
Im
Re
Open Question:How can we deter-mine u, v ∈ Rn andare they unique?
2. Placement in the Hermitian case
RCL circuits as Hermitian pencils
In the remainder, we will consider restricted perturbations ofthe form
sE − A + sxxT , x ∈ Rn
with additional assumption E = ET ∈ Rn×n andA = AT ∈ Rn×n.
A large class of networks that can be described with Hermitianpencils are RCL circuits which can be decribed by the pencils
sE−A = s
ACCATC 0 0
0 −L 00 0 0
−−ARR
−1ATR −AL −AV
−ATL 0 0
−ATV 0 0
with incidence matrices AL, AC , AR , AV of inductors,capacitances, resistors and voltage sources.
Rank one eigenvalue placement for Hermitian pencils
Given HermitianA(s) = sE − A withσ(A) =
..{λ1, . . . , λm}m(λ1), . . . ,m(λm)
and
pws. dist. µ1, . . . , µl ∈ CConstruct x ∈ Rn withP(s) = (αs − β)xxT s. t.σ(A+ P) = {µ1, . . . , µl} .
Previous Results: Golub ’73 (E = In, A = AT , α = 0, β = −1),Elhay, Golub, Ram ’03 (E > 0, A = AT ).
Structure of Hermitian matrix pencils
Theorem (Thompson ’76; Rodman,Lancaster ’05)
Let sE − A a Hermitian regular pencil with simple finiteeigenvalues, then there exist S ∈ Cn×n invertible such thatS∗(sE − A)S is block diagonal with blocks
λ ∈ R : ελ(s − λ), ελ ∈ {−1, 1}.
λ ∈ C+ :
(0 s − λ
s − λ 0
)∈ C2×2,
λ =∞ : ε(i)∞ (−1), ε(i)∞ ∈ {−1, 1}, i = 1, . . . , n − r
with r := deg det(sE − A).
Characteristic signs and small perturbations of eigenvalues
The figure below shows how the characteristic signs determine theeigenvalue movement for perturbations P(s) = sxxT :
8
0
-1 1
ε∞ 1
Characteristic signs and small perturbations of eigenvalues
The figure below shows how the characteristic signs determine theeigenvalue movement for perturbations P(s) = sxxT :
8
0
-1 1
ε∞ -1
Main result: Solution of Hermitian EPP
Theorem
Let A(s) = sE − A be Hermitian and regular and given pairwisedistinct µ1, . . . , µr+1 ∈ C \ (σ(A) ∪ {0}), 0 /∈ σ(A), then thereexists P(s) = s xxT with
σ(A+ P) = {µ1, . . . , µr+1} ∪ {λ ∈ σ(A) : gmA(λ) ≥ 2}.if and only if for all λ ∈ σ(A) ∩ R with characteristic sign ελ
ελ = sgn −∏
λ′ ∈ σ(A) \ {λ,∞} λ′(λ′ − λ)∏r+1
i=1 (µi − λ)µiand
sgn
∏λ ∈ σ(A) \ {∞} λ∏r+1
i=1 µi∈ {ε(1)∞ , . . . , ε(n−r)∞ }.
Example: The double low pass filter
sE−A = s
(0 0 0 00 1 0 00 0 10 00 0 0 0
)−(−1 1 0 −1
1 −1.1 0.1 00 0.1 −0.1 0−1 0 0 0
)
Using MATLAB one finds that
S =
( 0 0 0.6687 0.6688−0.9996 −0.0290 0 00.0092 −0.3161 0 0−0.9996 −0.0290 −1.0820 0.4133
)is transforms the pencil to Hermitian canonical form
S∗(sE−A)S =
( s+1.1009 0 0 00 s+0.0091 0 00 0 −1 00 0 0 1
), ε1.1009 = ε0.0091 = ε(2)∞ = −ε(1)∞ = 1.
Example: The double low pass filter
Given µ1, µ2, µ3 6= 0 ans assume that µ1 < µ2 < µ3 < 0 Thereforethe sign conditions are as follows:
1 = ε−1.1009 = sgn −−0.0091(−0.0091− (−1.1009))
µ1µ2µ3(µ1 − (−1.1009))(µ2 − (−1.1009))(µ3 − (−1.1009)).
From this and the condition on ε−0.0091 we see that
(µ1 + 1.1009)(µ2 + 1.1009)(µ3 + 1.1009) < 0
(µ1 + 0.0091)(µ2 + 0.0091)(µ3 + 0.0091) > 0
Therefore we have interlacing of the eigenvalues on the real line:
8 0
ε∞
-1.1009
ε-1.1009=1
8
-0.0091
μ1
ε-1.1009=1
μ2 μ3
-
What happens for non-simple eigenvalues
Consider A(s) =−(
0 s + 1s + 1 1
)⊕(
0 s − 1s − 1 1
), α = 1, β = 0.
We vary µ1 = −1 with amA+P(µ1) = 2 and display all possiblevalues for µ2 ∈ C \ R, µ3 = µ2, amA+P(µ2) = amA+P(µ3) = 1.
−10 −5 0 5 10−10
−8
−6
−4
−2
0
2
4
6
8
10
What happens for non-simple eigenvalues
Consider A(s) =−(
0 s + 1s + 1 1
)⊕(
0 s − 1s − 1 1
), α = 1, β = 0.
We vary µ1 = −3 with amA+P(µ1) = 2 and display all possiblevalues for µ2 ∈ C \ R, µ3 = µ2, amA+P(µ2) = amA+P(µ3) = 1.
−10 −5 0 5 10−10
−8
−6
−4
−2
0
2
4
6
8
10
What happens for non-simple eigenvalues
Consider A(s) =−(
0 s + 1s + 1 1
)⊕(
0 s − 1s − 1 1
), α = 1, β = 0.
We vary µ1 = −3.5 with amA+P(µ1) = 2 and display all possiblevalues for µ2 ∈ C \ R, µ3 = µ2, amA+P(µ2) = amA+P(µ3) = 1.
−10 −5 0 5 10−10
−8
−6
−4
−2
0
2
4
6
8
10
What happens for non-simple eigenvalues
Consider A(s) =−(
0 s + 1s + 1 1
)⊕(
0 s − 1s − 1 1
), α = 1, β = 0.
We vary µ1 = −5 with amA+P(µ1) = 2 and display all possiblevalues for µ2 ∈ C \ R, µ3 = µ2, amA+P(µ2) = amA+P(µ3) = 1.
−10 −5 0 5 10−10
−8
−6
−4
−2
0
2
4
6
8
10
What happens for non-simple eigenvalues
Consider A(s) =−(
0 s + 1s + 1 1
)⊕(
0 s − 1s − 1 1
), α = 1, β = 0.
We vary µ1 = −7 with amA+P(µ1) = 2 and display all possiblevalues for µ2 ∈ C \ R, µ3 = µ2, amA+P(µ2) = amA+P(µ3) = 1.
−10 −5 0 5 10−10
−8
−6
−4
−2
0
2
4
6
8
10
3. The end of the saga?
How to use the results
We assume that sE − A is Hermitian and consider the restrictedperturbations of the form
Pij(s) = scij(ei − ej)(ei − ej)T . (1)
Choose a desired eigenvalue configuration {µ1, . . . , µr+1}then two cases can occur:
1. Case: The sign conditions ελ = ... are not satisfied. Thenthere is no Hermitian perturbation P(s) = sxxT . Hence thereis no restricted perturbation (1).2. Case: The sign conditions are satisfied. In this case thecorresponding perturbation P(s) = sxxT is (up to the sign)uniquely determined.
More on the 2. Case
Let S∗(sE − A)S be in Hermitian canonical form and assume thatthe sign conditions are satisfied then P(s) = sxxT is given withSx = y = (yλ1 , . . . , yλr , y
T∞)T by
yλi = ±
√√√√ ∏λ′ ∈ σ(A) \ {λi ,∞}
|λ′||λi − λ′|
r+1∏j=1
|λi − µj ||µj |
, λi ∈ R
y2λi = −i∏
λ′ ∈ σ(A) \ {λ,∞}
−λ′
λ− λ′r+1∏j=1
λ− µj−µj
, yλi = yλi , imλi > 0
‖y∞‖2 =
∏λ′∈σ(A)\{∞} |λ′|∏r+1
j=1 |µj |.
Then we simply minimize mini ,j mincij>0 ‖Sx −√cijS(ei − ej)‖ and
the minimizer is√cij =
(Sx)∗S(ei−ej )‖S(ei−ej )‖2
.
Error bound in the chordal distance
We identify points z ∈ C ∪ {∞} with points (1, z)T on theRiemann sphere and introduce the chordal distance as
χ(
(γδ
),
(γ
δ
)) :=
|γδ − δγ|√|γ|2 + |δ|2
√|γ|2 + |δ|2
.
Then for every (γ, δ)T ∈ σ(A+ Pij) there exists(γ, δ)T ∈ σ(A+ P) such that
χ(
(γδ
),
(γ
δ
)) ≤
√2√‖Sx‖2 + ‖dij‖2‖S‖‖S−1‖
min{σmin([E + uuT , A]), σmin([E + dijdTij , A])}
‖Sx −√cijS(ei − ej)‖
with E = S∗ES , E = S∗ES , dij :=√cijS(ei − ej).
To be continued...
Non-Hermitian matrix pencils
In the non-Hermitian case we have the following placementconditions. Assume that there exist S ,T ∈ Cn×n such that
S(sE − A)T =
[sIr − diag (λi )
ri=1 0
0 −In−r
]with λi 6= 0 and define
Su := (v1, . . . , vr , vT∞)T , TTu := (u1, . . . , ur , u
T∞)T .
Given pairwise distinct µ1, . . . , µr+1 ∈ C \ {0} then there exists aperturbation P(s) = suuT satisfyingσ(A+ P) \ {∞} = {µ1, . . . , µr+1} if and only if there is a solutionof
viui =
∏rj=1 λj
∏r+1j=1 (λi − µj)
λi (−µr+1)∏r
j=1 µj(λi − λj), vT∞u∞ =
∏ri=1 λi∏r+1i=1 µi
.