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

.