Minimization of the Frobenius norm of a complex matrix using planar similarities

Applied Numerical Mathematics 40 (2002) 391–414www.elsevier.com/locate/apnum

Minimization of the Frobenius norm of a complex matrixusing planar similarities

Richard GabrielHardtstrasse 36, D-76185 Karlsruhe, Germany

Abstract

We investigate one step of anoptimal norm-reducing method with shearTkj on the pivot pair(k, j) :A �→TkjAT

−1jk without the usual restriction det(Tkj ) = 1 [Eberlein, SIAM J. Appl. Math. 10 (1962) 74–88;

Paardekooper, Doctoral Dissertation, Eindhoven Institute of Technology, 1969]. The normh(Z) = ‖TkjAT −1kj ‖F

can be expressed in terms of the nontrivial elements ofX = Z∗Z. HereZ is the essential part of the shearTkj .The resolvent and structural parameters that appear by solving the problem inf{h(Z), detZ �= 0} are invariant withrespect to some unitary shear transformations. This study providesinvariant formulasfor Paardekooper’s methodthat is based on the solution of the problem inf{h(Z), detZ = 1}. 2001 IMACS. Published by Elsevier ScienceB.V. All rights reserved.

1. Introduction

The aim of this paper is to find an algebraical solution to the problem

inf∥∥TkjAT −1

kj

∥∥2, detTkj = detZ �= 0. (∗)

We apply three transformations to the objective function in order to get thecanonical form(∗∗) be-low. For solving the problem, we have to distinguish between an interior optimum and an (authentic)infimum. From Section 3 until Section 11, the infimum is assumed to be a minimum. By Lagrange’s the-orem, we obtain two polynomial equations in the variables det(Tkj )

2 (asx) and the Lagrange parameter(asy), respectively; we shall call these polynomialsresolvents. Motivated by the “structural theorems”,the coefficients of these polynomials will be calledstructural parameters.

Beside Theorem 4, dealing with invariance of structural parameters under some complex planar rota-tions, the main contributions of this paper are Theorem 1 concerning the feasibility area for the solutionof the resolvent and Theorem 9 concerning the uniqueness of the solution in this area.

In this paper, the canonical form of the(k, j)-cross is appropriate only for a theoretical study. Forpractical purposes, e.g., for a Jacobi-like algorithm, the formulas given in Lemma 3 should be used for

E-mail address:[email protected] (R. Gabriel).

0168-9274/01/$22.00 2001 IMACS. Published by Elsevier Science B.V. All rights reserved.PII: S0168-9274(01)00092-7

392 R. Gabriel / Applied Numerical Mathematics 40 (2002) 391–414

structural parameters and the formula given in Theorem 3 should be used for the solution matrixX. Thisstudy enables us to solve Paardekooper’s problem with invariant formulas.

Numerical examples illustrate the two algorithms provided in the paper.

2. The objective function

Let A ∈ Cn×n be a complex matrix. Aplanar similarity is induced by a matrix of the form:

Tkj =

1 . . . 0 . . . 0 . . . 0...

......

...

0 . . . zkk . . . zkj . . . 0...

......

...

0 . . . zjk . . . zjj . . . 0...

......

...

0 . . . 0 . . . 0 . . . 1

.

We will assume only that detZ �= 0, where

Z =[zkk zkjzjk zjj

]is the “active” part of the matrix. A similarityTkjAT

−1kj acts only on the(k, j)-crossof the matrix, i.e.,

only on the union of rows and columnsk andj . Let us denote:

Akj :=[ak

aj

]= rowsk andj of A lessA

[k, j

k, j

]=

(akk akjajk ajj

),

Akj := [ak,aj

] = columnsk andj of A lessA[k, j

k, j

],

Ajk := the “rest” of the matrix.

Further, let

U :=AkjA∗kj =

[‖ak‖2 aka∗j

aja∗k ‖aj‖2

]:=

[a1 c1

c1 b1

],

V := (Akj

)∗Akj =

[ ‖ak‖2 (ak)∗aj

(aj )∗ak ‖aj‖2

]:=

[a1 c1

c 1 b1

],

W :=A

[k, j

k, j

]− t

2E, t = akk + ajj .

The objective function for problem(∗) has initially the form∥∥TkjAT −1kj

∥∥2 = ‖ZAkj‖2 +∥∥∥∥ZA[

k, j

k, j

]Z−1

∥∥∥∥2

+ ∥∥AkjZ−1∥∥2 + ∥∥Ajk

∥∥2

= Tr(Z∗ZAkjA

∗kj

) + Tr(Z∗ZA

[k, j

k, j

](Z∗Z)−1A∗

[k, j

k, j

])+ Tr

[(Akj

)∗Akj

(Z∗Z

)−1] + ∥∥Ajk

∥∥2.

R. Gabriel / Applied Numerical Mathematics 40 (2002) 391–414 393

Now, we shall apply three transformations to the variable matrixZ and to the(k, y)-cross aiming toobtain the simplest form for the objective function.

(a) First, putZ∗Z =X.(b) ForX use the form

X = κ

[x1 z

z x2

]with x1x2 − |z|2 = 1.

(c) After a(k, j) planar rotation withakk = ajj , the(k, j)-cross achieves a canonical form.Now, if we neglect the constant terms‖Aj

k‖2 and 2|akk|2, the objective function takes the form

h(X) = h(x1, x2, z, κ)

= |akj |2x21 + |ajk|2x2

2 − ajk akj z2 − akj ajk z 2

+ κ(a1x1 + b1x2 + c1 z+ c1z

) + κ−1(a1x2 + b1x1 − c1 z− c 1z).

We shall pay attention especially to the optimization problem

inf h(x1, x2, z, κ), (x1, x2, κ) > 0, x1x2 − |z|2 = 1. (∗∗)

Once we have found a solution

X = κ

[x1 z

z x2

]it is easy to build a matrixZ (unique, except for a unitary factor) satisfyingZ∗Z = X. For example, ifZ is taken to be upper triangular, then

Z = √κ

1√x2

0

z√x2

√x2

. (1)

3. The case of an interior optimum

First, assume the optimization problem is not degenerate, i.e., the infimum is actually a minimum.Then the optimal solution lies in the set(x1, x2, κ) > 0. We write

Lh(x1, x2, z, κ, λ)= h(x1, x2, z, κ)+ λ(x1x2 − |z|2 − 1

)for the Lagrange function with multiplierλ and we cancel the derivatives with respect tox1, x2, z= ξ + iηandκ . We get a first system inx1 andx2:

2|akj |2x1 + λx2 = −κa1 − κ−1b1,(2)

λx1 + 2|ajk|2x2 = −κb1 − κ−1a1

and a second one inz and z:λz+ 2akj ajk z= κc1 − κ−1c1,

2 akj ajkz+ λ z= κ c1 − κ−1 c 1.


The two systems have the same determinantδ = λ2 − 4|akj ajk|2 (except for the sign) and, ifδ �= 0,then we get the solution

x1 = 1√xδ

[2|ajk|2(a1x + b1) − y

(b1x + a1)],

x2 = 1√xδ

[2|akj |2(b1x + a1) − y

(a1x + b1)], (3)

z = 1√xδ

[2akj ajk( c 1 − c1x

) − y(c1 − c1x

)],

wherex := κ2 andy = λ. The discriminant∆ of the characteristic polynomial of the matrix

A

[k, j

k, j

]is∆= 4akj ajk . If g := |akj ajk|, theng =∆/4 and it is an invariant for every(k, j)-similarity. Using thisnotation, the determinant becomesδ = λ2 − 4g2 = y2 − 4g2.

Finally, by cancelling the derivative with respect toκ , we get

∂L∂κ

= a1x1 + b1x2 + c1 z+ c1z− 1

κ2

(a1x2 + b1x1 − c 1z− c1 z ) = 0. (4)

4. The resolvents

If we substitute the solutions (3) in Eq. (4) and inx1x2 − |z|2 = 1, then we get two resolvents whoseintersection point leads just to the solution of problem(∗∗). First, let us define thestructural parametersfor a (k, j)-cross in canonical form as

f1 := a1b1 − ∣∣c1∣∣2,f 1 := a1b1 − ∣∣c1∣∣2,h1 := a1a

1 + b1b1 + c1 c 1 + c1c

1,(5)

d1 := |ajk|2a21 + |akj |2b2

1 − akj ajk c 21 − akj ajkc2

1,

d1 := |ajk|2(b1)2 + |akj |2(a1)2 − akj ajk( c 1)2 − akj ajk(c1)2,

h1 := |akj |2a1b1 + |ajk|2a1b1 + akj ajk c1 c 1 + akj ajkc1c

1.

The invariantg is regarded as a structural parameter, too.Substituting (3) in (4) we get the third degree resolvent

g(x, y) = x2(f1y − d1)− (f 1y − d1) = 0, (6)

and by substituting (3) inx1x2 − |z|2 = 1, a fifth degree resolvent is obtained. This can be put either inthe form

f (x, y) = −x(y2 − 4g2)2 + y2R(x)− 2yS(x) + 4g2R(x)= 0, (7)


where

R(x)= f1x2 + h1x + f 1,

S(x) = d1x2 + 2h1x + d1,

or in the form

f (x, y) = F(y)x2 − xG(y) +H(y)= 0 (7∗)

with

F(y) = f1y2 − 2d1y + 4f1g

2,

G(y) = (y2 − 4g2)2 − h1y

2 + 4h1y − 4h1g2,

H(y)= f 1y2 − 2d1y + 4f 1g2.

It is noteworthy that the coefficients of both resolvents are structural parameters and this is true for theoptimal value of the objective function, too: if formula (3) is used forx1 andx2, it is converted into

h(x1, x2, z, κ)=m= −2y3R(x)+ (3y2 − 4g2)S(x)

x(y2 − 4g2)2.

Now, taking (7) into account, the minimal value of the objective function can be expressed as

m= −S(x)+ 2xy2

x(y2 − 4g2). (8)

A possibility to solve the problem(∗∗) would be to replace the value ofy given by the third degreeresolvent, namely,

y = d1x2 − d1

f1x2 − f 1,

into the fifth degree resolvent; thus an equation of degree 10 is obtained. Among its 10 roots, we wouldhave to select that one which confers a minimal value tom given by (8). However, we are able to indicatean interval where only the solution(x, y) is located. We will depict the resolvents in this interval and wewill ascertain the solution.

5. Degeneracy

Problem (2) is degenerate if the determinantδ(λ) = λ2 − 4g2 vanishes. We have to distinguish twocases:

CaseI. λ= 2g. In this case system (2) takes the form

2|akj |2x1 + 2|akj ajk|x2 = −κa1 − κ−1b1, 2|akj ajk|x1 + 2|ajk|2x2 = −κb1 − κ−1a1.

Because the left-hand side is non-negative, while the right-hand side is non-positive, the following resultholds:

Lemma 1. The conditionsak = 0 = aj and ak = 0 = aj as well asakj = 0 = ajk are necessary andsufficient for aλ= 2g degeneracy.


CaseII. λ= −2g. In this case the system (2) takes the form

2|akj |2x1 − 2gx2 = −κa1 − κ−1b1, −2gx1 + 2|ajk|2x2 = −κb1 − κ−1a1.

The compatibility conditions are

2|akj ajk|(κb1 + κ−1a1) + 2|ajk|2(κa1 + κ−1b1) = 0,

2|akj |2(κb1 + κ−1a1) + 2|akj ajk|(κa1 + κ−1b1) = 0.

Since every term in the left-hand sides is non-negative, it is possible to devise various degeneracy criteriaas, for example:

Lemma 2. The conditionsa1 = 0 = b1 and a1 = 0 = b1 are necessary and sufficient for aλ = −2gdegeneracy.

Remark. In both cases, the eigenvalues of the matrix

A

[k, j

k, j

]are also eigenvalues of the matrixA.

6. The feasibility area of solution

Now, we examine what conditions should be imposed to the solution pair(x, y) if we take into accountthe constraintsx1 > 0, x2 > 0 of problem(∗∗). For this, we have to match the signs of the hyperbolas

X1(x, y) = 2|ajk|2(a1x + b1) − y(b1x + a1) = 0,

X2(x, y) = 2|akj |2(b1x + a1) − y(a1x + b1) = 0

in relationship with the bandδ(y) = y2 − 4g2. To this end, we will construct a table of signs. If werepresent the hyperbolas in the form

y1 = 2|ajk|2a1x + b1

b1x + a1and y2 = 2|akj |2b1x + a1

a1x + b1,

then it is easy to see thatx > 0 implies y > 0 and thaty1y2 = 4|ajkakj | = 4g2 is always true. Theasymptotes parallel to theOx axis are

y1 = 2|ajk|2a1

b1and y2 = 2|akj |2b1

a1,

while those parallel to theOy axis, which lay in the negative area, are

x1 = −a1

b1and x2 = −b1

a1.

Further, forx > 0, we have

limy→∞X1(x, y) = −∞ and lim

y→∞X2(x, y) = −∞.

It follows that the functionsX1(x, y) andX2(x, y) are positive under the hyperbolas and negative above.


Let X1(x, y) = max(X1(x, y),X2(x, y)) and X2(x, y) = min(X1(x, y),X2(x, y)). The followingholds true: Letx = x0 > 0 be a straight line parallel to theOy axis. The ordinatesy1 and y2 of itsintersection points with the hyperbolas are positive and are separated by the valuey = 2g. Finally, wetakex1 = X1/δ andx2 = X2/δ. Thus, the sign table below can be drawn:

y > y1 y1 � y > 2g 2g > y � y2 y2 > y >−2g −2g > y

X1(x0, y) − + + + +X2(x0, y) − − − + +δ(y) + + − − +x1 − + − − +x2 − − + − +

Hence we can state the following result:

Theorem 1. The conditions(i) x1 > 0, x2 > 0, x > 0,(ii) (x, y) ∈ (x > 0, y <−2g)

are equivalent.

Thus, the domain(x > 0, y <−2g) describes the feasible area for the solution pair(x, y).

7. The structural parameters are non-negative and invariant

Next, we will show that the structural parameters can be decomposed in non-negative constituentswhich can be expressed by invariant formulas.

Theorem 2. The structural parametersd1, d1, h1 and h1 can be decomposed into non-negative

summands as:d1 = (|ajk|a1 − |akj |b1

)2 + 2gf1 + 2[g|c1|2 − Re

(akj ajk c 2

1

)],

d1 = (|akj |a1 − |ajk|b1)2 + 2gf 1 + 2[g∣∣c1∣∣2 − Re

(akj ajk( c 1)2)]

,

h1 =(√

a1a1 −√b1b1

)2 + 2(√

a1a1b1b1 − ∣∣c1 c 1∣∣) + 2[∣∣c1 c 1∣∣ + Re

(c1 c 1)],

h1 =(|akj |

√a1b1 − |ajk|

√a1b1

)2 + 2g(√

a1a1b1b1 − ∣∣c1c1∣∣)

+ 2[∣∣akj ajkc1c

1∣∣ + Re( akj ajkc1c

1)].Proof. There is no difficulty in seeing that the splitting is correct. Concerning non-negativity: the firstterms are squares of real numbers, for the middle ones Cauchy’s theorem could be invoked, while for thelast ones we can apply the inequality|a| ± Re(a) > 0, valid for everya ∈ C. ✷


The matricesU, V andW have been defined in Section 2. For these matrices we have

U+ =[b1 −c1

− c1 a1

],

W ∗U+W =[

0 ajk akj 0

][b1 −c1

− c1 a1

] [0 akj

ajk 0

]=

[ |ajk|2a1 −ajk akj c1

−akj akj c1 |akj |2b1,

]and also

W ∗VW =[

0 ajk akj 0

][a1 c1

c 1 b1

][0 akj

ajk 0

]=

[ |ajk|2b1 akj ajk c 1

akj ajkc1 |akj |2a1

].

Consequently, we can state

Theorem 3. The solutions given in(3) can be expressed in matrix form as follows:

X = 1

y2 − 4g2

(−xyU+ − yV + 2xW ∗U+W + 2W ∗VW).

Lemma 3. The structural parameters can be expressed as:

f1 = detU, f 1 = detV, h1 = TrUV,

d1 = Tr(WUW ∗U+)

, d1 = Tr(WV +W ∗V

), h1 = Tr

(WUW ∗V

).

Theorem 4. The structural parameters remain invariant after a planar complex(k, j)-rotation.

Proof. As a result of a planar complex(k, j)-rotation, the(k, j)-cross(Akj ,A

[k, j

k, j

],Akj

)takes the form(

T Akj , T A

[k, j

k, j

]T ∗,AkjT ∗

).

Thus, the matricesU, V andW change into matricesU, V andW by the formulas

U = TAkj (T Akj )∗ = TAkjA

∗kj T

∗ = TUT ∗,V = (

AkjT ∗)∗(AkjT ∗) = T

(Akj

)∗AkjT ∗ = T V T ∗,

W = TA

[k, j

k, j

]T ∗ − t

2E = TWT ∗.

Further, if detU �= 0, then

U+ = (detU

)U−1 = (

detU)TU−1T ∗ = (detU)TU−1T ∗ = TU+T ∗

and this is true generally for the transpose matrix of algebraical complementsU+. Analogously,V + = T V +T ∗. Now, we have

f1 = detU = detT UT ∗ = detU = f1,

f 1 = detV = detT V T ∗ = detV = f 1,


h1 = Tr U V = Tr(TUT ∗)(T V T ∗) = Tr

(TUV T ∗) = TrUV = h1,

h1 = Tr(W UW ∗V

) = Tr(TWT ∗)(T UT ∗)(TW ∗T ∗)(T V T ∗)

= Tr(TWUW ∗V T ∗) = TrWUW ∗V = h1.

The proof ford1 andd1 is similar. ✷Theorem 5. After a planar complex(k, j)-rotation, the solution matrix takes the formX = TXT ∗.

Proof. Since the structural parameters remain invariant, the solution pair(x, y) and resolvents areinvariant too. We have

X = 1

y2 − 4g2

(−xyU+ − yV + 2xW ∗U+W + 2W ∗V W)

= 1

y2 − 4g2

(−xyT U+T ∗ − yT V T ∗ + 2xT(W ∗U+W

)T ∗ + 2T

(W ∗VW

)T ∗) = T XT ∗. ✷

8. Structural theorems

Here we will justify the termstructural parameters. Let

Z = d

[x11 x12

x21 x22

]with x11x22 − x12x21 = 1 andd �= 0.

Let us look closely how a(k, j)-similarity acts upon a(k, j)-cross which has the canonical formakk = ajj . We have

ZAkj = d

[x11ak + x12aj

x21ak + x22aj

], ZA

[k, j

k, j

]Z−1 =

[ ∗ akj x211 − ajkx

212

ajkx222 − akj x

221 ∗

],

AkjZ−1 = d−1 [ x22ak − x21a

j , −x12ak + x11a

j ] .

In Theorem 2 we have split structural parameters into three non-negative summands; hence canceling astructural parameter leads to three equations. Actually, we will see that there exists a deeper relationship.Let us examine

d1 = (|ajk|a1 − |akj |b1)2 + 2gf1 + 2

[g|c1|2 − Re

(akj ajk c 2

1

)]and f1 = a1b1 − |c1|2.

Theorem 6. If A ∈ Cn×n is a complex matrix, then for a(k, j)-cross of the matrix, the next two assertionsare equivalent:

(i) d1 = 0 = f1.(ii) There exists a(k, j)-planar similarity which brings the(k, j)-cross to the form

∗ 0 0

∗ ∗ aj

ak aj


except for a permutation. At least an eigenvalue of the matrix

A

[k, j

k, j

]is also an eigenvalue of the matrixA.

Proof. We assume that the(k, j)-cross is already reduced to the canonical formakk = ajj by using aplanar complex(k, j)-rotation which keeps (i) and (ii) valid. Condition (ii) is then formally equivalent tothe following condition: the system

xak + yaj = 0, (a)

x2akj − y2ajk = 0 (b)

has a non-trivial solution(x, y) �= 0. Obviously,f1 = 0 is equivalent to (a).Now, we consider the case where there is no degeneracy and alsoa1 �= 0 �= b1, which implies

x �= 0 �= y. Assume (i) is true. It follows fromd1 = 0 that

|ajk|a1 − |akj |b1 = 0, akj ajk c 21 − ∣∣akj ajkc2

1

∣∣ = 0.

Let w = x/y; then aj = −wak becausef1 = 0. Further,c1 = aka∗j = − wa1, henceakj ajk c 2

1 =w2akj ajka2

1 = |w2akj ajk|a21. It follows thatw2akj |ajk|2 − ajk|w2akj ajk| = 0, which gives

w2akj − ajk|w2akj ajk|

|ajk|2 = w2akj − ajk∣∣w2∣∣∣∣∣∣akjajk

∣∣∣∣= x2

y2akj − akj

∣∣∣∣x2

y2

∣∣∣∣∣∣∣∣akjajk

∣∣∣∣ = 1

y2

(x2akj − ajky

2∣∣∣∣x2

y2

∣∣∣∣∣∣∣∣akjajk

∣∣∣∣) = 0.

Now, fromaj = −wak we getb1 = |w2|a1. Then

|ajk|a1 − |akj |b1 = |ajk|a1 − |akj |∣∣w2∣∣a1 = 0

leads to the relation

|w|2 =∣∣∣∣xy

∣∣∣∣ =∣∣∣∣ajkakj

∣∣∣∣which givesx2akj − y2ajk = 0, and therefore (i) implies (iii).

Now assume (iii) holds. From (a) we haveb1 = |w2|a1 and c1 = −wa1, while from (b) we getajk = w2akj . Therefore,|ajk|a1 − |akj |b1 = |w2||akj |a1 − |akj ||w|2a1 = 0, akj ajk c 2

1 = akj w 2 akjw2a1 =|w|4|akj |2a1, which is real and non-negative, and it is easy to see that (iii) implies (i).

In case of degeneracy, the proof is simpler.✷Now, let us take a look at

h1 =(|akj |

√a1b1 − |ajk|

√a1b

1)2 + 2g

(√a1b1a

1b1 − ∣∣c1c1∣∣)

+ 2[∣∣akj ajkc1c

1∣∣ + Re( akj ajkc1c

1)],h1 =

(√a1a1 −

√b1b1

)2 + 2(√

a1a1b1b1 − c1 c 1)

+ 2[∣∣c1 c 1∣∣ + Re

(c1 c 1)].


Theorem 7. If A ∈ Cn×n is a complex matrix, then assuming that no special degeneracy occurs, for a(k, j)-cross of the matrix the next assertions are equivalent:

(i) h1 = 0 = h1.(ii) There exists a(k, j)-planar similarity which brings the(k, j)-cross to the form

∗ 0 0

∗ ∗ aj

ak 0

except for a permutation. Both eigenvalues of the matrix

A

[k, j

k, j

]are also eigenvalues of the matrixA.

Proof. We assume that the(k, j)-cross is already reduced to the canonical formakk = ajj by using aplanar complex(k, j)-rotation which keeps (i) and (ii) valid. Condition (ii) is then formally equivalent tothe following condition: the system

xak + yaj = 0, (a)

xaj − yak = 0, (b)

x2akj − y2ajk = 0 (c)

has a non-trivial solution(x, y) �= 0.We assume that (iii) holds and that no degeneracy is present. Let us denotew = −y/x. Then

aj = 1

wak, aj = −wak, akj =w2ajk.

Further,

b1 = 1

|w|2a1, b1 = |w|2a1, c1 = 1

wa1, c1 = −wa1.

We have√a1a1 −

√b1b1 =

√a1a1 −

√1

|w|2a1|w|2a1 = 0,

c1 c 1 = 1

wa1(− wa1) = −a1a

1 (real and non-positive),

|akj |√a1b1 − |ajk|

√a1b1 = |w|2|ajk|

√a1

1

|w|2a1 − |ajk|√a1|w|2a1 = 0,

akj ajkc1c1 = w 2ajk ajk 1

wa1(−w)a1 = −|ajk|2a1a1|w|2 (non-positive).


It follows thath1 = 0 = h1.Assume now that (i) is true. Fromh1 = 0 we geta1a

1b1b1 = |c1|2|c1|2. Hencea1b1 = |c1|2, a1b1 =

|c1|2. These givef1 = 0 = f 1, which impliesaj = uak andaj = νak so thatc1 = ua1 andc1 = νa1. Wecancel now the summands in Theorem 2 to get

|akj |√a1b1 − |ajk|

√a1b

1 = |akj |√a1|u|2a1 − |ajk|

√a1|ν|2a1 =

√a1a

1(|akj ||u| − |ajk||ν|) = 0.

It follows that |ν|/|u| = |akj |/|ajk| and, further,√a1a1 −

√b1b1 =

√a1a1 −

√|νu|2a1a1 = 0.

Now, a1a1(1− |νu|)= 0, which implies|ν||u| = 1. Further,∣∣c1 c 1∣∣ + c1 c1 = |u|a1|ν|a1 + ua1 νa1 = a1a

1(|νu| + ν u ) = 0.

Henceν = −1/u. It is essential to prove thatakju2 − ajk = 0 holds true. To achieve this aim we shall usethe third component ofh1. We have

akj ajku ν + ∣∣akj ajku ν∣∣ = 0.

We replaceν = −1/u to get

−akj ajk u u +∣∣∣∣akj ajk u u

∣∣∣∣ = 0.

Now, we multiply withajk , multiply with u and simplify with|ajk|2 to get

−akju2

|u|2 + ajk|akj ajku2||ajk|2|u|2 = 0.

Hence

akju2 − ajk

|akj ||ajk| |u|2 = 0.

But |akj |/|ajk| = ν/u= 1/|u|2 and this gives

akju2 − ajk = 0.

Now, it is obvious that condition (i) implies (iii). ✷In the same way, a result based on the conditiond1 = 0= f 1 can be formulated.

Corollary 1. If A ∈ Cn×n is a complex matrix, then for a(k, j)-cross of the matrix the followingassertions are equivalent:

(i) f1 = d1 = 0 = f 1 = d1, h1 + h1 �= 0.(ii) There exists a(k, j)-planar similarity which brings the(k, j)-cross to the form

∗ 0 0

0 ∗ aj

0 aj


except for a permutation. At least an eigenvalue of the matrix

A

[k, j

k, j

]is also an eigenvalue of the matrixA.

Remark. Let T be the essential part of the planar complex rotation which makesakk = ajj and let

Z =[ c1 −a1

−a1 c 1

],

where a1 and c1 are computed usingTAkj and a1, c1 using AkjT ∗, respectively. Then the planarsimilarity whose essential part isZ0 = ZT reduces the order of matrixA with at least one unity inall of the three cases.

9. On the geometry of resolvents

(a) The resolvent of third degree can be expressed

either asy = d1x2 − d1

f1x2 − f 1or asx = ±

√f 1y − d1

f1y − d1.

We havedy

dx= 2x

D

(f1x2 − f 1)2,

where the discriminantD = f1d1 − f 1d1 is exhibited. The graph has two vertical asymptotesx =√

f 1/f1, x = −√f 1/f1 and a horizontal oney = d1/f1. The vertical axisOy is a symmetry axis for the

third degree resolvent. It is obvious that the graph has three branches and that only one half of a branchlies in the feasible domain.

In case of degeneracy,D = 0, we get(f 1y − d1

)(f1x

2 − f 1) = 0,

and the graph degenerates into asymptotes; only a vertical one(x = √f 1/f1) intersects the feasible area.

(b) The fifth degree resolvent is an equation of second degree in thex variable; this fact enables us toexpressx as function ofy:

x1,2 = G(y)± √G2(y)− 4F(y)H(y)

2F(y).

According to the sign of the discriminant∆=G2(y)−4F(y)H(y), a horizontal line intersects the graphin two or no points; on the other hand, a vertical line intersects the graph either in two or in four points.

Lemma 4. If no special degeneracy occurs, the horizontal asymptotes lie in the upper half-plane.

Proof. The horizontal asymptotes are given by the roots of the equation

F(y) = f1y2 − 2d1y + 4f1g

2 = 0.


Fig. 1.

The discriminant of this equation is

D1 = 4d21 − 16g2f 2

1 = 4(d1 − 2gf1)(d1 + 2gf1)� 0.

The non-negativity follows from Theorem 2. The roots of the equation satisfy the conditions

y1 + y2 = 2d1, y1y2 = 4f1g2, y1, y2 real.

Thus,y1, y2 � 0. ✷A full analysis of the geometry of the fifth degree resolvent exceeds the limits of this paper.

Nevertheless, in Fig. 1 a typical situation forn � 4 is sketched. Besides, we will completely examinethe behavior of the graph in the feasible area.

Eq. (7∗) can be written in the form

x− = 2H(y)

G(y)+ √∆(y)

, x+ = 2H(y)

G(y)− √∆(y)

.

Hence

limy→∞x− = 0, lim

y→∞x+ = +∞.


There are no other vertical, horizontal or oblique asymptotes. This shows also that there exists at least apartial branch of the graph lying in the domain(x > 0, y <−2g). We will prove that there exists exactlyone.

Lemma 5. If no special degeneracy occurs, then no branch of the graph of the fifth degree resolventcrosses boundaries of the feasibility domain of the solution pair(x, y).

Proof. For x = 0 we get

H(y)= f 1y2 − 2d1y + 4f 1g2 = 0.

For the same reasons as in the proof of Lemma 4, we havey1, y2 � 0. Further, ify = −2g , then from (6)we obtain

4g2R(x)+ 4gS(x)+ 4g2R(x)= 4g[2gR(x)+ S(x)

] = 0.

If g = 0, then the axisy = 0 belongs to the graph of the resolvent; ifg �= 0, then Eq. (7) becomes

x2(2gf1 + d1)+ 2x(gh1 + h1) + 2gf 1 + d1 = 0.

This equation has a positive root only in the case of degeneracy:

f1 = f 1 = h1 = 0 = h1 = d1 = d1. ✷

10. There exists a unique branch of the fifth degree resolvent in the feasibility area

In order to prove the uniqueness, we shall show that every vertical linex = x0 cuts the fifth degreeresolvent in exactly one point in the feasibility domain(x > 0, y <−2g). To this end, let us examine thefamily of (k, j)-crosses(

µAkj ,A

[k, j

k, j

],µAkj

), µ ∈ R. (9)

The corresponding family of fifth degree resolvents is

f (x, y,µ) = −x(y2 − 4g2)2 +µ4[R(x)y2 − 2yS(x) + 4g2R(x)

] = 0.

Settingµ4 = 1/s, we get the equation

s = R(x)y2 − 2yS(x) + 4g2R(x)

x(y2 − 4g2)2. (10)

Obviously, the mappingy → s is injective.

Theorem 8. If no ordinary degeneracy occurs, then in the feasible area lies exactly one branch of thefifth degree resolvent.

Proof. We will also show that the mappings → y is injective. Let us have a look at the derivative

∂s

∂y= 2[−y3R(x)+ 3y2S(x)− 12g2yR(x)+ 4g2S(x)]

x(y2 − 4g2)3.


For x > 0, y < −2g we have∂s/∂y > 0. The equation∂s/∂y = 0 is equivalent toh1 = f1 = f 1 =0 = h1 = d1 = d1. But a continuous function with a positive derivative is injective and this proves thetheorem. ✷

11. There exists a unique intersection point of the resolvents in the feasibility area

Let us consider again the family (9) of(k, j)-crosses and the corresponding family (10) of fifth degreeresolvents. Notice that the third degree resolvents remain unchanged. Hence we have only to prove thatthe mappings → y induced by the intersection point(x, y, s), is injective.

Lemma 6. We haved1d1 � g|D|.

Proof. From Theorem 2 we getd1 = 2gf1 + p1, d1 = 2gf 1 + p1 with p1 � 0, p1 � 0. We distinguish

two cases.(1) D � 0. This meansf1d

1 � f 1d1. Hence

g|D| � gf1d1 � 2gf1d

1 � d1d1.

(2) D � 0. Thus

g|D| � gf 1d1 � 2gf 1d1 � d1d1.

Theorem 9. If no ordinary degeneracy occurs then the third and fifth degree resolvents intersect in thefeasible area in exactly one point.

Proof. It suffices to show that the derivativeds

dy= ∂s

∂y+ ∂s

∂x

dx

dy

is positive in the feasible area. In the above formula, dx/dy is the derivative of the third degree resolvent.Let us compute it at an intersection point:

ds

dy= 2

x(y2 − 4g2)3

[R(x)

(−y3 − 12g2y) + S(x)

(3y2 + 4g2) − D2(y2 − 4g2)2

4(d1 − f 1y)(d1 − f1y)2

].

Now, fromx > 0, y <−2g we have 2/x(y2 − 4g2)3 > 0. Inequalities (i) and (ii) below are equivalent inthe feasible areax > 0, y <−2g:

−R(x)(y3 + 12g2y

) + S(x)(3y2 + 4g2) > D2(y2 − 4g2)2

4(d1 − f 1y)(d1 − f1y)2, (i)

−f 1(y3 + 12g2y) + d1(3y2 + 4g2) > D2(y2 − 4g2)2

4(d1 − f 1y)(d1 − f1y)2. (ii)

We obtain (ii) if we takex = 0 in (i), while (i) results from (ii) when we add to both sides a positivequantity, namely,

−(f1x

2 + h1x)(y3 + 12g2y

) + (d1x

2 + 2h1x)(

3y2 + 4g2).


The denominator is positive fory <−2g, so that (ii) gives

4(d1 − f 1y

)(d1 − f1y)

2[−f 1(y3 + 12g2y) + d1(3y2 + 4g2)] �D2(y2 − 4g2)2

.

Computing the left-hand side, we get:

L = 4f 21

(f 1)2

y6 − [16f 2

1 f1d1 + 8d1f1

(f 1)2]

y5

+ 4[d2

1

(f 1)2 + 8d1d

1f1f1 + 3f 2

1

(d1)2 + 12f 2

1

(f 1)2

g2]y4

− 4[4d2

1d1f 1 + 6d1f1

(d1)2 + 4g2f 2

1 d1f 1 + 12g2f 1(f 2

1 d1 + 2d1f1f

1)]y3

+ 4[3d2

1

(d1)2 + 12g2d2

1

(f 1)2 + 24g2d1d

1f1f1 + 4g2f 2

1

(f 1)2 + 8g2d1d

1f1f1]y2

− 4[12g2f 1d1d2

1 + 4g2d21d

1f 1 + 8g2(d1)2d1f1

]y + 16g2d2

1

(d1)2

.

For the right-hand side we obtain:

R =D2y4 − 8d2g2y2 + l6D2g4.

It is manifest that all terms inL are non-negative fory < 0. It remains only to show that all terms inLare greater than the corresponding terms inR. We have

a4 = 4(d2

1

(f 1)2 + 8d1d

1f1f1 + 3f 2

1

(d1)2 + 12f 2

1

(f 1)2

g2)� D2 = f 2

1

(d1)2 − 2f1d

1f 1d1 + (f 1)2

d21 .

Actually, the difference is

a4 −D2 = 11f 21

(d1)2 + 10d1d

1f1f1 + 3d2

1

(f 1)2 + 48f 2

1

(f 1)2

g2 � 0.

We have

a0 = 16g2d21

(d1)2 � 16D2g4.

The non-negativity of the difference

a0 − 16D2g4 = 16g2[d21

(d1)2 −D2g2]

is a direct consequence of Lemma 6.

Remark. Problem(∗∗) typically has exactly one solution. A closer examination of the intersection ofresolvents in the feasible domain enables us to give better limits for the cutting pointS; namely, ifD < 0,then √

2gf 1 + d1

2gf1 + d1� x �

√f 1

f1

while the inequalities are reversed ifD > 0 (see Fig. 2).


Fig. 2.

12. On the case of an authentical infimum in problem (∗∗)

In order to study this situation, let us remark that the objective functionh(x1, x2, z, κ) in problem(∗∗)consists originally of three non-negative components

h(x1, x2, z, κ)=E1(x1, x2, z)+ κE2(x1, x2, z)+ κ−1E3(x1, x2, z),

where

E1(x1, x2, z)= |akj |2x21 + |ajk|2x2

2 − ajk akj z2 − akj ajk z 2,

E2(x1, x2, z)= a1x1 + b1x2 + c1z+ c1 z,E3(x1, x2, z)= a1x2 + b1x1 − c 1z− c1 z.

In turn, these non-negative components can be further divided into smaller non-negative components:

Ei(x1, x2, z)=Ei1(x1, x2)+Ei2(x1, x2, z), i = 1,2,3,

where

E11(x1, x2)= (|akj |x1 − |ajk|x2)2 � 0,

E12(x1, x2, z)= 2[gx1x2 − Re

(akj ajkz2)]

� 2(gx1x2 − ∣∣akj ajk∣∣|z|2) = 2g

(x1x2 − |z|2) = 2g � 0,

E21(x1, x2)= (√a1

√x1 − √

b1√x2

)2 � 0,


E22(x1, x2, z)= 2[√

a1b1√x1x2 + Re

( c1z)]

� 2(√

a1b1√x1x2 − ∣∣ c1z

)= 2

(√a1b1

√1+ |z|2 − ∣∣ c1

∣∣|z|) � 2(√

a1b1|z| −∣∣ c1

∣∣|z|) = 2|z|(√a1b1 − |c1|) � 0,

E31(x1, x2)=(√

b1√x1 −

√a1

√x2

)2� 0,

E32(x1, x2, z)= 2(√

a1b1√x1x2 + Re

( c 1z))

� 0.

It is not possible to treat in a unitary manner the situation of an authentical infimum. Common is onlythe fact that at infimum, the functionsEi as well asE1s , κE2s andκ−1E3s are finite, even when one ormore variables go to infinity. For the casesκ → ∞ andκ−1 → ∞ (κ > 0) it is important to compute theglobal infimum of the functionE2 andE3.

Lemma 7. We have

inf E2(x1, x2, z)= 2√f1, x1x2 − |z|2 = 1,

inf E3(x1, x2, z)= 2√f 1, x1x2 − |z|2 = 1.

Proof ( for E2). The corresponding Lagrange function forE2 is

L(x1, x2, z, t)= a1x1 + b1x2 + c1 z+ c1z− t(x1x2 − |z|2 − 1

).

The derivatives with respect tox1 andx2 are

∂L∂x1

= a1 − tx2 = 0,∂L∂x2

= b1 − tx1 = 0,

while those with respect toz= x + iy are

∂L∂x

= c1 + c1 + 2tx,∂L∂y

= −ic1 + i c1 + 2ty.

Hence

∂L∂x

+ i∂L∂y

= 2c1 + 2tz = 0.

We obtain

x2 = a1

t, x1 = b1

t, z = −c1

t.

In case of degeneracy,t = 0, we havea1 = 0 = b1 (c1 = 0) which impliesAkj = 0. We can computet inthe caset �= 0:

x1x2 − |z|2 − 1 = 1

t2a1b1 − |c1|2 1

t2− 1 = 1

t2f1 − 1 = 0.

Thereforet = √f1. Hence

inf E2(x1, x2, z)= 1

t

(2a1b1 − 2|c1|2) = 2

√f1. ✷


Lemma 8. In the case of an authentical infimum induced byκ → ∞, a necessary condition isf1 = 0,while for one induced byκ → 0 a necessary condition isf 1 = 0.

Let us denotez = |z|eiθ andakj ajk = geiϕ.

Theorem 10. In the case of an authentical infimum induced byx1

x2→ a ∈ R+, x1, x2 → ∞, κ → κ0 ∈ R+, θ → θ0,

the conditionsf1 = 0 = f 1 andd1 = 0= d1 are necessary.

Proof. We have:

E11 = x22

(|akj |x1

x2− |ajk|

)2

→ ∞(|akj |a − |ajk|)2,

E21 = x2

(√a1

√x1

x2− √

b1

)2

→ ∞(√a1

√a − √

b1)2,

E31 = x2

(√b1

√x1

x2− √

a1

)2

→ ∞(√

b1√a − √

a1)2.

By canceling the parentheses, we get

a = |ajk||akj | = a1

b1= b1

a1.

Further, it follows fromx1x2 − |z|2 = 1 that∣∣∣∣ zx2

∣∣∣∣2 = x1

x2− 1

x22

→ a.

Thusz2/x22 → ae2iθ0 holds. Hence

E12 = 2x22

{gx1

x2− Re

[akj ajk

(z

x2

)2]}→ ∞ga

(1− Ree2i(θ−θ0)

),

E22 = 2x2

[√a1b1

√x1

x2+ Re

( c1

z

x2

)]→ ∞√

a[√

a1b1 + Re( c1eiθ0

)],

E32 = 2x2

[√a1b1

√x1

x2+ Re

( c 1 z

x2

)]→ √

a[√

a1b1 + Re(c1eiθ0

)].

Canceling again the parentheses, we get

e2iθ0 = akj ajkg

= a1b1

c 21

= a1b1

( c 1)2.

If we replace now these expressions in the suitable terms of the equations appearing in Theorem 2, weobtaind1 = 0= d1 as well asf1 = 0= f 1. ✷


In the same way it is possible to prove the following results.

Lemma 9. In the case of an authentical infimum induced byx1/x2 → ∞, x1, x2 → ∞ the conditionsg = 0, a1 = b1 = 0 = a1 = b1 andakj + ajk �= 0 are necessary.

Lemma 10. In the case of an authentical infimum induced byx1/x2 → ∞, x2 → a ∈ R+ the conditionsakj = 0, a1 = 0= b1 are necessary.

Remark. If more structural parameters are null as in Theorem 10 for example, we have to use structuraltheorems.

13. Invariant formulas for Paardekooper’s method

The problem

inf∥∥TkjAT −1

kj

∥∥, detTkj = detZ = 1 (∗.1)

was solved for the first time by Paardekooper [8]. Actually, if we takeκ = 1 in our method, weobtain invariant formulas for this problem, which are quite different from those of Paardekooper. Letus summarize. Starting with a(k, j)-cross(

Akj ,A

[k, j

k, j

],Akj

)of a complex matrixA ∈ C

n×n, we build the 2× 2 matricesU,V,W as in Section 2 and the structuralparametersf1, f

1, d1, d1, h1, h

1 as described in Lemma 3. We denoteg = |∆|/4, with ∆ = t2 − 4d,d = akkajj − akj ajk , andz= akk + ajj .

Now, if in the fifth degree resolventf (x, y) = 0 we makex = 1, then we get a fourth degree resolvent:

f (1, y) = −(y2 − 4g2)2 + (

y2 + 4g2)(f1 + h1 + f 1) − 2y(d1 + 2h1 + d1) = 0.

Using the unique solutiony <−2g, we compute the Hermitian matrix

X = 1

y2 − 4g2

[−y(U+ + V

) + 2W ∗(U+ + V)W

] =[x1 z

z x2

],

where we use the notation[a b

c d

]+=

[d −b

−c a

]and, finally, the solution matrix

Z =

1√x2

0

z√x2

√x2

which solves problem(∗.1).


14. On Jacobi-like methods

We use the notationΓ R-κ for the norm-reducing method based on the solution of problem(∗), whilefor the Paardekooper method operating with our formulas as described in Section 13, the notationPΓ -1is reserved. We will compare the two methods and locate their place among other Jacobi-like methods.We think there are three classes of norm-reducing Jacobi-like methods:

(a) Methods which operate with simple planar similarities which are rooted in [1]. To this familybelong the method of Huang and Gregory [14], the method of Voevodin and Ikramov [13], as wellas theΓ R-κ andPΓ -1 methods.

(b) Methods operating with special triangular matrices; they were introduced by Veselic [12].(c) Methods developed for parallel computers, operating with a direct product of planar similarities.

This family includes the method of Sameh [11] and the method of Paardekooper [9]. The last oneis an optimal single-step algorithm.

An Jacobi-like method acts normally in two phases, but in special cases a third phase is needed:(i) A norm-reducing similarity used to decrease the matrix norm.(ii) A planar complex rotation used to increase the amount of the matrix main diagonal.(iii) A perturbing similarity used if the process converges to a non-diagonal matrix (see Goldstein and

Horowitz [6] or Voevodin and Ikramov [13]).As to (ii), we now expose an explicit method. LetA,U ∈ C

2×2 and consider the problemmax‖DiagUAU ∗‖, UU ∗ = E. Let s = a11 + a22, d = detA and ∆ = s2 − 4d. We construct theHermitian matrixR = ∆ 1/2A + ∆1/2A∗ or R = A + A∗ if ∆ = 0. Let R = SΛS∗ be the canonicaldecomposition. ThenU = S∗ is the solution of the maximum problem.

Other trigonometric formulas for (ii) are given by Ruhe [10] and Voevodin and Ikramov [13].The methodΓ R-κ is an optimal single-step algorithm and there is no difficulty to find numerically the

intersection point of the resolvents. The methodPΓ -1 is an optimal unimodular single-step algorithm.

15. Minimization of the Euclidean norm

We now compare the methodsΓ R-κ andPΓ -1. Since the latter is obtained from the former puttingκ = 1, the single-steps for theΓ R-κ method are better. The question is how is this reflected in the wholealgorithm. It is hard to find a theoretical answer, so we carried out many numerical experiments. First, letus see how both methods work on a matrix of order 4. Let, for example,

A =

1+ i 2 − i 2 + 3i 4− 5i

1− i 2 + 4i 1− i 3 − 2i

5+ 6i 1− i 5 − 2i 4+ i

2− 4i 3− 2i 4+ 5i 3+ 2i

.

From Tables 1 and 2 it can be seen that pivots are chosen cyclically. In Table 1 we represent the normof the matrix and that of the diagonal, at each iteration, for the methodΓ R-κ . In Table 2, the same datafor the methodPΓ -1 are given.

Both methods converge to the diagonal matrix

D = diag(11.408928+ i1.7032990,1.3854683+ i4.7118307,

0.1011842− i7.8343722,−1.8955819+ i6.4192425).


Table 1

It. k j Norm Norm Dg

1 2 3 16.580886 8.559727

2 1 2 16.332694 10.467209

3 2 4 16.284246 12.549437

4 1 3 16.259360 12.682206

5 1 2 16.254676 15.668772

6 2 4 16.250260 15.700098

7 2 3 16.246842 16.050805

8 2 4 16.231343 16.117152

9 1 3 16.230317 16.192958

10 1 2 16.229332 16.223541

11 1 4 6.228864 16.227120

12 2 3 16.228836 16.227249

13 3 4 16.228832 16.228787

14 2 4 16.228779 16.228779

15 1 3 16.228779 16.228779

16 2 3 16.228779 16.228779

17 1 2 16.228779 16.228779

18 1 4 16.228779 16.228779

19 3 4 16.228779 16.228779

Table 2

It. k j Norm Norm Dg

1 1 2 17.059727 8.136127

2 2 3 16.701849 8.640068

3 3 4 16.610473 8.741935

4 1 3 16.587578 11.820505

5 2 4 16.543957 11.931242

6 1 4 16.539253 12.226251

7 1 2 16.474197 15.850682

8 2 3 16.353557 15.881659

9 3 4 16.347239 16.051503

10 1 3 16.295386 16.089078

11 2 4 16.294605 16.090740

12 1 4 16.237490 16.224967

13 1 2 16.230316 16.226757

14 2 3 16.230155 16.228800

15 3 4 16.229572 16.228805

16 1 3 16.229570 16.228805

17 2 4 16.228780 16.228772

18 1 4 16.228780 16.228772

19 1 2 16.228779 16.228779

In order to reach a precision of 10−6, the methodΓ R-κ needs 14 steps, while for the methodPΓ -1,19 iterations are needed.

Generally speaking, based on our numerical experiments with various matrices up to order 300, wecan say that theΓR-κ method is on average 5% quicker than thePΓ -1 method. We explain the smalldifference, despite of the extra parameter, by the fact that in theΓ R-κ method the parameterκ isapproaching unity as the number of iterations increases.

Acknowledgements

I would like to express my thanks to L. Elsner, M. Iosifescu and M.H.C. Paardekooper for their remarksconcerning a previous version of this paper.


References

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[2] A. Erdmann, Über Jacobi-ähnliche Verfahren zur Lösung des Eigenwetproblems nichtnormaler Matrizen,Dissertation, Fernuniversität Hagen, 1984.

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Minimization of the Frobenius norm of a complex matrix using planar similarities