Diagonal entries of Hadamard Product of a Totally Non ...³2.pdf · Totally Nonnegative Matrices, series in applied mathematics 2011 Maite Gass o Diagonal entries of Hadamard Product

IntroductionProblem

Diagonal entries of Hadamard Product of a TotallyNon Positive Matrix and its inverse

Maite Gasso

Institut de Matematica Multidisciplinar,Universitat Politecnica de Valencia, Valencia, Spain

Based in joint work with R. Bru, I.Gimenez and M. Santana

Jornadas sobre matrices totalmente positivas y totalmente negativas,CIEM, Castro-Urdiales,

5–6 Marzo, 2015.

Maite Gasso Diagonal entries of Hadamard Product of a Totally Non Positive Matrix and its inverse

IntroductionProblem

Index

1 IntroductionHadamard ProductRelative Gain Array A ◦ A−T

2 ProblemDiagonal entries

Diagonal entries: Symmetric CaseDiagonal entries: General Case


IntroductionProblem

Index





IntroductionProblem

Hadamard ProductRelative Gain Array A ◦ A−T

Index





IntroductionProblem


A ◦ B is the Hadamard (entrywise) product: (A ◦ B)ij = aijbij

trigonometric moments of convolutions of periodic functions

products of integral equation kernels (relationship with Mercer’s theorem)

characteristic functions in probability theory (Bochner’s theorem)

operator theory the Hadamard product for infinite matrices

etc

Markham A semigroup of totally nonnegative matrices Linear Algebra andAppl 1970

Johnson. Clousure Properties of certain positivity classes of matrices.Linear Algebra and Appl 1987

Jurgen Garloff, David G. Wagner. Preservation of Total Nonnegativityunder the Hadamard Product and Related Topics . Total Positivity and ItsApplications. Mathematics and Its Applications Volume 359, 1996, pp 97-102

Johnson and Fallat. Totally Nonnegative Matrices, series in appliedmathematics 2011


IntroductionProblem


A ◦ B is the Hadamard (entrywise) product: (A ◦ B)ij = aijbij

trigonometric moments of convolutions of periodic functions

products of integral equation kernels (relationship with Mercer’s theorem)

characteristic functions in probability theory (Bochner’s theorem)

operator theory the Hadamard product for infinite matrices

etc

Markham A semigroup of totally nonnegative matrices Linear Algebra andAppl 1970

Johnson. Clousure Properties of certain positivity classes of matrices.Linear Algebra and Appl 1987

Jurgen Garloff, David G. Wagner. Preservation of Total Nonnegativityunder the Hadamard Product and Related Topics . Total Positivity and ItsApplications. Mathematics and Its Applications Volume 359, 1996, pp 97-102

Johnson and Fallat. Totally Nonnegative Matrices, series in appliedmathematics 2011


IntroductionProblem


Relative Gain Array A ◦ A−T


IntroductionProblem




IntroductionProblem




IntroductionProblem



The combined matrix of a nonsingular matrix A:

C (A) = A ◦(A−1

)T= A ◦ A−T

If we multiply A by a non singular diagonal matrix from the left orfrom the right, C (A) will not change.

C (A) = [cij ] →∑j

cij = 1 ∀i and∑i

cij = 1 ∀j

C (A) ≥ 0 ⇒ C (A) is a doubly stochastic matrix

R. Bru, M.T. Gasso, I. Gimenez and M. Santana. NonnegativeCombined Matrices. Journal of Applied Mat. Advances in Matrices. 2014.


IntroductionProblem




C (A) = A ◦(A−1

)T= A ◦ A−T


C (A) = [cij ] →∑j


cij = 1 ∀j




IntroductionProblem




C (A) = A ◦(A−1

)T= A ◦ A−T


C (A) = [cij ] →∑j


cij = 1 ∀j




IntroductionProblem




C (A) = A ◦(A−1

)T= A ◦ A−T


C (A) = [cij ] →∑j


cij = 1 ∀j




IntroductionProblem


C (A) = A ◦(A−1

)T= A ◦ A−T

Relationship between diagonal entries and eigenvalues of adiagonalizable matrix:

B = [bij ] diagonalizable with eigenvalues λ1, λ2, . . . , λn

⇒ exists nonsingular A such that B = A[diag(λ1, λ2, . . . , λn)]A−1

⇒ C (A)

λ1

λ2

...λn

=

b11

b22

...bnn

R. Horn and C.R. Johnson, Topics in Matrix Analysis, CambridgeUniversity Press, 1991.


IntroductionProblem


C (A) = A ◦(A−1

)T= A ◦ A−T




⇒ C (A)

λ1

λ2

...λn

=

b11

b22

...bnn



IntroductionProblem


C (A) = A ◦(A−1

)T= A ◦ A−T




⇒ C (A)

λ1

λ2

...λn

=

b11

b22

...bnn



IntroductionProblem


C (A) = A ◦(A−1

)T= A ◦ A−T




⇒ C (A)

λ1

λ2

...λn

=

b11

b22

...bnn



IntroductionProblem


C (A) = A ◦(A−1

)T= A ◦ A−T




⇒ C (A)

λ1

λ2

...λn

=

b11

b22

...bnn



IntroductionProblem

Diagonal entries

Index





IntroductionProblem

Diagonal entries

Motivation

It is long standing problem to characterize the range of C (A) if A runsthrough the set of all positive definite nxn matrices. A partial answerdescribing the diagonal entries of C (A) is the case positive definitematrix case given by Fiedler in 1964.

small M. Fiedler. Relations between the diagonal entries of twomutually inverse positive definite matrices. Czechosl. Math.J.14(89), 1964, 39− 51.


IntroductionProblem

Diagonal entries

Motivation


IntroductionProblem

Diagonal entries

Recently, Fiedler and Markham tried to characterize the properties of thesequence of the diagonal entries uij of the combined matrix C (A) for anytotally positive matrix A.

Theorem

If A = [aij ] be an n × n STP matrix, if n ≥ 3, the combined matrix C (A)satisfy

u11 < u22,

un−1,n−1 > unn

as well as

uii > 1 for all i .

M. Fiedler and L. Markham. Combined matrices in special classes ofmatrices. Linear Algebra and its Applications. 435: 1945-1955, 2011.


IntroductionProblem

Diagonal entries

Recently, Fiedler and Markham tried to characterize the properties of thesequence of the diagonal entries uij of the combined matrix C (A) for anytotally positive matrix A.

Theorem

If A = [aij ] be an n × n STP matrix, if n ≥ 3, the combined matrix C (A)satisfy

u11 < u22,

un−1,n−1 > unn

as well as

uii > 1 for all i .

M. Fiedler and L. Markham. Combined matrices in special classes ofmatrices. Linear Algebra and its Applications. 435: 1945-1955, 2011.


IntroductionProblem

Diagonal entries

Problem

Problem 1

Find necessary and sufficient conditions for the sequence of the diagonalentries of C (A) if A is totally negative n × n matrices.

Problem 2

Characterize the range of C (A) if A runs through the set of all totallynegative n × n matrices.


IntroductionProblem

Diagonal entries

Problem

Problem 1

Find necessary and sufficient conditions for the sequence of the diagonalentries of C (A) if A is totally negative n × n matrices.

Problem 2

Characterize the range of C (A) if A runs through the set of all totallynegative n × n matrices.


IntroductionProblem

Diagonal entries

Previous Results

Theorem

If A is an invertible TN–matrix, then SA−1S is strictly sign regular withsignature ε = (1, . . . , 1,−1).

S.Fallat. On matrices with all minors nonnegative. Electronic LinearAlgebra, 92–99, 2000

Theorem

The necessary and sufficient condition for a real matrix A = [−aik ],aik > 0 to be totally negative is that all relevant submatrices be totallynegative

R.Canto, A.Urbano.B. Ricarte Some properties about totally nonpositive matrix ELA 2010


IntroductionProblem

Diagonal entries

Diagonal entries

Lemma 1

Let A = [−aij ], i = 1, . . . , n − 1, k = 1, . . . , n, be a totallynegative matrix, if n ≥ 3, and −a11 = −a12, then

det

−a11 −a13 · · · −a1n

−a21 −a23 · · · −a21

...... · · ·

...−an−1,1 −an−1,3 · · · −an−1,n

< det

−a12 −a13 · · · −a1n

−a22 −a23 · · · −a21

...... · · ·

...−an−1,2 −an−1,3 · · · −an−1,n

(1)


IntroductionProblem

Diagonal entries

Sketch of the proof

We use induction on n, if n = 3, the matrix A has the form

A =

[−a11 −a12 −a13

−a21 −a22 −a23

].

As A is totally negative, det

[−a11 −a12

−a21 −a22

]< 0 and (a11 = a12) implies

a22 < a21 and indeed

det

[−a11 −a13

−a21 −a23

]< det

[−a12 −a13

−a22 −a23

].


IntroductionProblem

Diagonal entries

For n = 3, the lemma is true. Suppose now that n > 3 and that forn − 1the result holds. Let B be the matrix which coincides with A in all entriesexcept for an−1,n, and such that its entry −an−1,n satisfies

det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

= 0 (2)

and−an−1,n = (−an−1,n)− ε1, ε1 > 0 (3)

For matrix B holds

det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

= 0 ≥ det

−a11 −a13 −a1n

−a21 −a23 −a2n

......

...−an−1,1 −an−1,3 −an−1,n

< 0


IntroductionProblem

Diagonal entries

Suppose now that n > 3 and that for (n − 1) the result holds.

det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

= 0 > det

−a11 −a13 −a1n

−a21 −a23 −a2n

......

...−an−1,1 −an−1,3 −an−1,n

.

Note that for any ε > 0, the matrix B =

−a11 −a12 −a1n

−a21 −a22 −a2n

......

...−an−1,1 −an−1,2 −an−1,n + ε

is totally negative since B and A have the same system of relevant submatricesexcept that on the left-hand since .


IntroductionProblem

Diagonal entries

By induction hypothesis, the cofactor C1 of entry an−1,n on the left-hand sideof (1) is less than the corresponding cofactor C2 of entry an−1,n in the rightside-hand side of (1),

C1 = det

−a11 −a13 −a1n−1

−a21 −a23 −a2n−1

......

...−an−2,1 −an−2,3 −an−2,n−1

< C2 = det

−a12 −a13 −a1n−1

−a22 −a23 −a2n−1

......

...−an−2,2 −an−2,3 −an−2,n−1

.

The left-hand side of (1) can be written as

det

−a11 −a13 −a1n

−a21 −a23 −a2n

......

...−an−1,1 −an−1,3 −an−1,n − an−1,n + an−1,n

=

(−an−1,n + an−1,n)C1 + det

−a11 −a13 −a1n

−a21 −a23 −a2n

......

...−an−1,1 −an−1,3 −an−1,n

.


IntroductionProblem

Diagonal entries

The right-hand side of (1) can be written as

det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n − an−1,n + an−1,n

=

(−an−1,n + an−1,n)C2 + det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

.

det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

= 0. (4)

Since by (4), the last determinant is 0, the inequalities C1 < C2 , ε > 0 andan−1,n > an−1,n imply that

det

−a11 −a13 −a1n

−a21 −a23 −a2n

......

...−an−1,1 −an−1,3 −an−1,n

< det

−a12 −a13 −a1n

−a22 −a23 −a2n

......

...−an−1,2 −an−1,3 −an−1,n

.Maite Gasso Diagonal entries of Hadamard Product of a Totally Non Positive Matrix and its inverse

IntroductionProblem

Diagonal entries

Diagonal Entries

Theorem 1

Let A = [−aij ] be an n × n totally negative matrix and let A−1 = [αij ]. Ifn ≥ 3, then the diagonal entries uii = [−aii ][αii ] of the combined matrixC (A) verifies

u11 > u22,

un−1,n−1 < unn.

If n = 2 then u11 = u22. as well as uii < 0 for all i .


IntroductionProblem

Diagonal entries

The combined matrix C(A) of a matrix A does not change if we multiply a row or acolumn by a positive number. So we can thus assume that a11 = 1, a12 = 1, a22 = 1and our problem is then to show that in the partitioning of A as

A =

−1 −1 −A13

−a21 −1 −A23

−A31 −A32 −A33

det

(−1 −A13

−A31 −A33

)< det

(−1 −A23

−A32 −A33

)By the lemma 1 removing the second row from A.

A′ = A[1, 3|1, 2, 3] =

(−1 −1 −A13

−A31 −A32 −A33

)Applying the lemma 1 to A′

det

(−1 −A13

−A31 −A33

)< det

(−1 −A13

−A32 −A33

)(5)

Using the lemma for columns after removing from A the first column, we obtain


IntroductionProblem

Diagonal entries

Diagonal Entries

Corollary for Totally Nonpositive Matrices

Let A = [−aij ] be an n × n t.n.p. nonsingular matrix, let A−1 = [αij ]. Ifn ≥ 3, then the diagonal entries uii = [−aii ][αii ] of the combined matrixC (A) verifies

u11 ≥ u22.

un−1,n−1 ≤ unn

as well asuii < 0 for all i .

Example

A =

0 −1 −2 −3−15 −15 −15 −15−15 −14 −12 −8−30 −27 −20 −3

→ C(A) =

0 . . .. −105 . .. . −108 .. . . −1.5


IntroductionProblem

Diagonal entries

Diagonal Entries

Corollary for Totally Nonpositive Matrices

Let A = [−aij ] be an n × n t.n.p. nonsingular matrix, let A−1 = [αij ]. Ifn ≥ 3, then the diagonal entries uii = [−aii ][αii ] of the combined matrixC (A) verifies

u11 ≥ u22.

un−1,n−1 ≤ unn

as well asuii < 0 for all i .

Example

A =

0 −1 −2 −3−15 −15 −15 −15−15 −14 −12 −8−30 −27 −20 −3

→ C(A) =

0 . . .. −105 . .. . −108 .. . . −1.5


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3× 3 symmetrictotally negative matrix

Lemma 2

A 3× 3 symmetric matrix is totally negative if and only if it is positivelydiagonally congruent to the matrix

A =

−1 −x3 −x2

−x3 −1 −x1

−x2 −x1 −1

where x1, x2, x3 are positive numbers satisfying x2 > x1x3, and thedeterminant, ∆ = −1− 2x1x2x3 + x2

1 + x22 + x2

3 , is negative.


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3× 3 symmetrictotally negative matrix

Theorem for Symmetric Totally Negative Matrices

A necessary and sufficient condition for three numbers u1 ,u2 and u3 tobe the diagonal entries of the combined matrix of a 3× 3 symmetrictotally negative matrix is that all ui be less than zero and

u1 + u3 − u2 − 1 > 0

Note that:If xi =

√1− ui , i = 1, 2, 3, the next matrix

A =

−1 −x3 −x2

−x3 −1 −x1

−x2 −x1 −1

is a t.n symmetric matrix. We have proved that with this conditions,x1.x2, x3 are positive numbers satisfying x2 > x1x3 and the determinant,∆ = −1− 2x1x2x3 + x2

1 + x22 + x2

3 < 0.


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3 × 3 symmetric t.n. matrix ( sufficient condition)

Example

Letu1 = −2, u2 = −8, u3 = −3

be three negative numbers that verifies the condition of the theorem,

u1 + u3 − u2 − 1 > 0.

We construct a symmetric matrix T , satisfying diag(C (T )) = (u1, u2, u3).

T =

−0.66443 −1 −1.00527−1 −1 −1

−1.1604 −1 −0.748111

and

C (T ) =

−2 4.92693 −1.926933.07307 −8 5.92693

−0.0730739 4.07307 −3

·Maite Gasso Diagonal entries of Hadamard Product of a Totally Non Positive Matrix and its inverse

IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3× 3 symmetric t.n. matrix (necessary condition)

Example

A =

−1 −3 −8−3 −1 −2−8 −2 −1

→ C(A) =

−0.15 1.95 −0.81.95 −3.25 2.2−0.8 2.2 −0.4

u1 = −0.15, u2 = −3.15, u3 = −0.4

(u1 + u3 − 1) = −1.55>− 3.15 = u2


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3× 3 totallynegative matrix

Lemma

A 3× 3 matrix is totally negative if and only if it is positively diagonallycongruent to the matrix

A =

− v1

v1 + 1−1 −p

−1 −1 −1

−q −1 − v3

v3 + 1

where v1, v3, p, q > 1 and ∆ =

(p − 1)(q − 1)(v1 + 1)(v3 + 1)− 1

(v1 + 1)(v3 + 1)< 0.


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of 3× 3 totallynegative matrix

Note

Observe that if A = [−aij ] is a t.n. regular matrix, then A=D1SD2, where

D1 =

1

a120 0

01

a220

0 01

a23

D2 =

a22

a210 0

0 1 0

0 0a22

a23

and

S =

−a11a22

a12a21−1 −a13a22

a12a23−1 −1 −1

−a31a22

a32a21−1 −a33a22

a32a23


IntroductionProblem

Diagonal entries

Theorem for totally negative Matrices General Case

A necessary and sufficient condition for three numbers, u1, u2, u3, to bethe diagonal entries of the combined matrix of a 3× 3 totally negativematrix is that all ui be less than zero and

u1 + u3 − u2 − 1 > 0


IntroductionProblem

Diagonal entries

We prove that the system:v1 = u1sv3 = u3spq(v1 + 1)(v3 + 1)− v1v3

s= u2

has solution with v1 > 0, v3 > 0, p > 1, q > 1 and s < 0.So we can define a matrix

A =

−

v1

v1 + 1−1 −p

−1 −1 −1

−q −1 −v3

v3 + 1

where u1, u2, u3 are the diagonal entries of the combined matrix.

Example

Letu1 = −2, u2 = −8, u3 = −3

be three negative numbers that verifies the condition of the theorem.

u1 + u3 − u2 − 1 > 0.

We construct a matrix T t.n such that diag(C (T )) = (u1, u2, u3).


IntroductionProblem

Diagonal entries

Example

T =

−0.66443 −1 −1.00527−1 −1 −1

−1.1604 −1 −0.748111

and

C (T ) =

−2 4.92693 −1.926933.07307 −8 5.92693

−0.0730739 4.07307 −3

·


IntroductionProblem

Diagonal entries

Theorem for totally negative Matrices General Case

A necessary and sufficient condition for three numbers, u1, u2, u3, to bethe diagonal entries of the combined matrix of a 3× 3 totally negativematrix is that all ui be less than zero and

u1 + u3 − u2 − 1 > 0

Example

A =

−8 −14 −19−13 −20 −27−17 −26 −35

→ C(A) =

−8−430

−385


IntroductionProblem

Diagonal entries

Diagonal Entries of totally negative Matrices

SUMARY

INEQUALITIES FOR t.n MATRICES

Lemma

A = [−aij ] t.n matrix, −a11 = −a12

⇓det(An2) < det(An1)

Theorem

A = [−aij ] A−1 = [αij ]

⇓[−a11][α11]>[−a22][α22]

[−an−1,n−1][αn−1,n−1]<[−ann][αnn]

[aii ][αii ]>0 for all i


IntroductionProblem

Diagonal entries

Diagonal Entries of totally negative Matrices

SUMARY

INEQUALITIES FOR t.n MATRICES

Lemma

A = [−aij ] t.n matrix, −a11 = −a12

⇓det(An2) < det(An1)

Theorem

A = [−aij ] A−1 = [αij ]

⇓[−a11][α11]>[−a22][α22]

[−an−1,n−1][αn−1,n−1]<[−ann][αnn]



IntroductionProblem

Diagonal entries

Diagonal Entries

INEQUALITIES FOR STP MATRICES AND t.n MATRICES

Theorem for STP (Fiedler 2011)

A = [aij ], A−1 = [αij ]

⇓[a11][α11]<[a22][α22]

[an−1,n−1][αn−1,n−1]>[ann][αnn]


Theorem for t.n.

A = [−aij ],A−1 = [αij ]

⇓[−a11][α11]>[−a22][α22]

[−an−1,n−1][αn−1,n−1]<[−ann][αnn]

−[aii ][αii ]<0 for all i


IntroductionProblem

Diagonal entries

Diagonal Entries

INEQUALITIES FOR STP MATRICES AND t.n MATRICES

Theorem for STP (Fiedler 2011)

A = [aij ], A−1 = [αij ]

⇓[a11][α11]<[a22][α22]

[an−1,n−1][αn−1,n−1]>[ann][αnn]


Theorem for t.n.

A = [−aij ],A−1 = [αij ]

⇓[−a11][α11]>[−a22][α22]

[−an−1,n−1][αn−1,n−1]<[−ann][αnn]

−[aii ][αii ]<0 for all i


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of Symetric 3× 3 Matrices

Conditions for an ordered n-tuple of real numbers to be ordered n-tupleof the diagonal entries of the combined matrix

Theorem for STP (F. and M.,2011)

u1,u2 and u3

to be diagonal entries of C(A)

⇐⇒ui>1

u1 + u3 − 1<u2≤(√

(u1) +√

(u3)− 1)2

Theorem for t.n.

the numbers u1,u2 and u3


⇐⇒ui<0

u1 + u3 − 1>u2


IntroductionProblem

Diagonal entries

Diagonal Entries of the Combined of Symetric 3× 3 Matrices

Conditions for an ordered n-tuple of real numbers to be ordered n-tupleof the diagonal entries of the combined matrix

Theorem for STP (F. and M.,2011)

u1,u2 and u3


⇐⇒ui>1

u1 + u3 − 1<u2≤(√

(u1) +√

(u3)− 1)2

Theorem for t.n.

the numbers u1,u2 and u3


⇐⇒ui<0

u1 + u3 − 1>u2


IntroductionProblem

Diagonal entries

THANKS


Documents

Diagonal entries of Hadamard Product of a Totally Non ...³2.pdf · Totally Nonnegative Matrices, series in applied mathematics 2011 Maite Gass o Diagonal entries of Hadamard Product