Transcript
Page 1: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

On a problem of Halmos:Unitary equivalence of a matrix to its transpose

Stephan Ramon Garcia

Pomona CollegeClaremont, California

March 18, 2011

Abstract

Halmos asked whether every square complex matrix is unitarily equivalent to its transpose (UET).Ad hoc examples indicate that the answer is no. In this talk, we give a complete characterizationof matrices which are UET. Surprisingly, the naıve conjecture that a matrix is UET if and only if itis unitarily equivalent to a complex symmetric (i.e., self-transpose) matrix is true in dimensionsn ≤ 7 but false for n ≥ 8. In particular, unexpected building blocks begin to appear in dimensions6 and 8. This is joint work with James E. Tener (UC Berkeley).

Partially supported by NSF Grant DMS-1001614 (SRG) and a NSF Graduate Research Fellowship (JET).

[email protected] http://pages.pomona.edu/~sg064747

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A “Halmos Problem”

Problem (Halmos, LAPB - Problem 159)

Is every square matrixunitarily equivalent to itstranspose (UET)?

Solution (Halmos)

Ad-hoc methods show that0 1 00 0 20 0 0

is not UET. Can we characterize those matrices which are UET?

[email protected] http://pages.pomona.edu/~sg064747

Page 3: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A “Halmos Problem”

Problem (Halmos, LAPB - Problem 159)

Is every square matrixunitarily equivalent to itstranspose (UET)?

Solution (Halmos)

Ad-hoc methods show that0 1 00 0 20 0 0

is not UET.

Can we characterize those matrices which are UET?

[email protected] http://pages.pomona.edu/~sg064747

Page 4: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A “Halmos Problem”

Problem (Halmos, LAPB - Problem 159)

Is every square matrixunitarily equivalent to itstranspose (UET)?

Solution (Halmos)

Ad-hoc methods show that0 1 00 0 20 0 0

is not UET. Can we characterize those matrices which are UET?

[email protected] http://pages.pomona.edu/~sg064747

Page 5: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 6: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 7: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 8: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 9: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 10: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Who Cares?

Why is this interesting?

It’s technically a “Halmos Problem”

Every square matrix is similar to its transpose, making theproblem somewhat difficult.

Linear preservers of the numerical range are of the formX 7→ UXU∗ or X 7→ UX tU∗ (where U is unitary). Thus wecharacterize the fixed points of linear preservers in thenontrivial case.

It has an obvious “answer,” which is totally wrong.

Funny things happen once you hit dimensions 6 and 8.

[email protected] http://pages.pomona.edu/~sg064747

Page 11: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A naive conjecture

Definition

If T is unitarily equivalent to a complex symmetric (i.e., self-transpose) matrix, then T is UECSM.

A Coincidence?

Halmos’ matrix 0 1 00 0 20 0 0

is the “simplest” example of a matrix which is not UECSM.

Naive Conjecture

UET ⇐⇒ UECSM (⇐ trivial)

[email protected] http://pages.pomona.edu/~sg064747

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A naive conjecture

Definition

If T is unitarily equivalent to a complex symmetric (i.e., self-transpose) matrix, then T is UECSM.

A Coincidence?

Halmos’ matrix 0 1 00 0 20 0 0

is the “simplest” example of a matrix which is not UECSM.

Naive Conjecture

UET ⇐⇒ UECSM (⇐ trivial)

[email protected] http://pages.pomona.edu/~sg064747

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A naive conjecture

Definition

If T is unitarily equivalent to a complex symmetric (i.e., self-transpose) matrix, then T is UECSM.

A Coincidence?

Halmos’ matrix 0 1 00 0 20 0 0

is the “simplest” example of a matrix which is not UECSM.

Naive Conjecture

UET ⇐⇒ UECSM (⇐ trivial)

[email protected] http://pages.pomona.edu/~sg064747

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UECSMs are Tricky

Exactly one of the following is UECSM:

(0 7 00 1 20 0 6

) (0 7 00 1 30 0 6

) (0 7 00 1 40 0 6

) (0 7 00 1 50 0 6

) (0 7 00 1 60 0 6

)

Which one is UECSM?

The fourth matrix is unitarily equivalent to2 +

√572

0 − 12i

√37− 73

√2

57

0 2−√

572

− 12i

√37 + 73

√2

57

− 12i

√37− 73

√2

57− 1

2i

√37 + 73

√2

573

[email protected] http://pages.pomona.edu/~sg064747

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UECSMs are Tricky

Exactly one of the following is UECSM:

(0 7 00 1 20 0 6

) (0 7 00 1 30 0 6

) (0 7 00 1 40 0 6

) (0 7 00 1 50 0 6

) (0 7 00 1 60 0 6

)

Which one is UECSM?

The fourth matrix is unitarily equivalent to2 +

√572

0 − 12i

√37− 73

√2

57

0 2−√

572

− 12i

√37 + 73

√2

57

− 12i

√37− 73

√2

57− 1

2i

√37 + 73

√2

573

[email protected] http://pages.pomona.edu/~sg064747

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UECSMs are Tricky

Exactly one of the following is UECSM:

(0 7 00 1 20 0 6

) (0 7 00 1 30 0 6

) (0 7 00 1 40 0 6

) (0 7 00 1 50 0 6

) (0 7 00 1 60 0 6

)

Which one is UECSM?

The fourth matrix is unitarily equivalent to2 +

√572

0 − 12i

√37− 73

√2

57

0 2−√

572

− 12i

√37 + 73

√2

57

− 12i

√37− 73

√2

57− 1

2i

√37 + 73

√2

573

[email protected] http://pages.pomona.edu/~sg064747

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A Gallery of UECSMs

3 6 60 4 10 0 3

∼= 1

2

(7−√

74)

3i

√2

109

(74−

√74)

12i

√1

109

(2629 + 72

√74)

3i

√2

109

(74−

√74)

3 + 36√

74109

3109

√5476 + 218

√74

12i

√1

109

(2629 + 72

√74)

3109

√5476 + 218

√74 1

218

(763 + 37

√74)

3 5 4

0 8 40 0 3

∼= 1

2

(11−

√82)

2i

√2

73

(82− 5

√82)

12i

√1

73

(1537 + 160

√82)

2i

√2

73

(82− 5

√82)

3 + 16√

8273

273

√6724 + 730

√82

12i

√1

73

(1537 + 160

√82)

273

√6724 + 730

√82 1

146

(803 + 41

√82)

5 2 2

7 0 07 0 0

∼= 1

2

(5 −√

187)

−5i

√561+5

√187

1658−i

√3350829− 125

√187

1658

−5i

√561+5

√187

16581

829

(1870 + 293

√187)

9829

√12

(173723 + 7075

√187)

−i

√3350829− 125

√187

16589

829

√12

(173723 + 7075

√187)

81−5+3

√187

9 8 9

0 7 00 0 7

∼= 8−

√1492

92 i

√16837+64

√149

13093 i

√13367213093 −

1296√

14913093

92 i

√16837+64

√149

13093207440+9477

√149

2618618√

3978002+82324√

14913093

i

√13367213093 −

1296√

14913093

18√

3978002+82324√

14913093

92675+1808√

14913093

[email protected] http://pages.pomona.edu/~sg064747

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Another Contender?

[email protected] http://pages.pomona.edu/~sg064747

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Skew-Hamiltonian Matrices

Definition

If Bt = −B and Dt = −D, then

T =

(A BD At

)is a skew-Hamiltonian matrix (SHM).

If T is unitarily equivalent toa SHM, then T is UESHM.

UESHM ⇒ UET

If T is skew-Hamiltonian, then T = JT tJ∗ where

J =

(0 I−I 0

).

[email protected] http://pages.pomona.edu/~sg064747

Page 20: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Skew-Hamiltonian Matrices

Definition

If Bt = −B and Dt = −D, then

T =

(A BD At

)is a skew-Hamiltonian matrix (SHM). If T is unitarily equivalent toa SHM, then T is UESHM.

UESHM ⇒ UET

If T is skew-Hamiltonian, then T = JT tJ∗ where

J =

(0 I−I 0

).

[email protected] http://pages.pomona.edu/~sg064747

Page 21: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Skew-Hamiltonian Matrices

Definition

If Bt = −B and Dt = −D, then

T =

(A BD At

)is a skew-Hamiltonian matrix (SHM). If T is unitarily equivalent toa SHM, then T is UESHM.

UESHM ⇒ UET

If T is skew-Hamiltonian, then T = JT tJ∗ where

J =

(0 I−I 0

).

[email protected] http://pages.pomona.edu/~sg064747

Page 22: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Reducibility of Skew-Hamiltonian Matrices

Theorem (Waterhouse, 2004)

If T is skew-Hamiltonian, then T is similar to A⊕ At for some A.

Proposition

If T is skew-Hamiltonian and 6× 6 or smaller, then T is reducible.

Pf. Sketch, 6× 6 case.

QT = TQ and Q = Q∗ lead to a 72× 36 real linear system,which is too large to consider symbolically.

Exploiting symmetries and using various reductions, oneobtains an equivalent 30× 21 real linear system.

Mathematica’s symbolic MatrixRank command shows thatthe rank is ≤ 20, whence a nonscalar Q exists.

[email protected] http://pages.pomona.edu/~sg064747

Page 23: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Reducibility of Skew-Hamiltonian Matrices

Theorem (Waterhouse, 2004)

If T is skew-Hamiltonian, then T is similar to A⊕ At for some A.

Proposition

If T is skew-Hamiltonian and 6× 6 or smaller, then T is reducible.

Pf. Sketch, 6× 6 case.

QT = TQ and Q = Q∗ lead to a 72× 36 real linear system,which is too large to consider symbolically.

Exploiting symmetries and using various reductions, oneobtains an equivalent 30× 21 real linear system.

Mathematica’s symbolic MatrixRank command shows thatthe rank is ≤ 20, whence a nonscalar Q exists.

[email protected] http://pages.pomona.edu/~sg064747

Page 24: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Reducibility of Skew-Hamiltonian Matrices

Theorem (Waterhouse, 2004)

If T is skew-Hamiltonian, then T is similar to A⊕ At for some A.

Proposition

If T is skew-Hamiltonian and 6× 6 or smaller, then T is reducible.

Pf. Sketch, 6× 6 case.

QT = TQ and Q = Q∗ lead to a 72× 36 real linear system,which is too large to consider symbolically.

Exploiting symmetries and using various reductions, oneobtains an equivalent 30× 21 real linear system.

Mathematica’s symbolic MatrixRank command shows thatthe rank is ≤ 20, whence a nonscalar Q exists.

[email protected] http://pages.pomona.edu/~sg064747

Page 25: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Reducibility of Skew-Hamiltonian Matrices

Theorem (Waterhouse, 2004)

If T is skew-Hamiltonian, then T is similar to A⊕ At for some A.

Proposition

If T is skew-Hamiltonian and 6× 6 or smaller, then T is reducible.

Pf. Sketch, 6× 6 case.

QT = TQ and Q = Q∗ lead to a 72× 36 real linear system,which is too large to consider symbolically.

Exploiting symmetries and using various reductions, oneobtains an equivalent 30× 21 real linear system.

Mathematica’s symbolic MatrixRank command shows thatthe rank is ≤ 20, whence a nonscalar Q exists.

[email protected] http://pages.pomona.edu/~sg064747

Page 26: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Reducibility of Skew-Hamiltonian Matrices

Theorem (Waterhouse, 2004)

If T is skew-Hamiltonian, then T is similar to A⊕ At for some A.

Proposition

If T is skew-Hamiltonian and 6× 6 or smaller, then T is reducible.

Pf. Sketch, 6× 6 case.

QT = TQ and Q = Q∗ lead to a 72× 36 real linear system,which is too large to consider symbolically.

Exploiting symmetries and using various reductions, oneobtains an equivalent 30× 21 real linear system.

Mathematica’s symbolic MatrixRank command shows thatthe rank is ≤ 20, whence a nonscalar Q exists.

[email protected] http://pages.pomona.edu/~sg064747

Page 27: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Irreducible skew-Hamiltonians exist if n ≥ 8

Numerical Evidence

Computation indicates that generic 8× 8 SHMs are irreducible.Finding a scalable family of provably irreducible SHMs is difficult.

Proposition

For d ≥ 4, the matrix T =

(A B0 A

), where

A =

1

2. . .

d

, B =

0 1 1 · · · 1 1

−1 0 1. . . 1 1

−1 −1 0. . . 1 1

.... . .

. . .. . .

. . ....

−1 −1 −1. . . 0 1

−1 −1 −1 · · · −1 0

,

is skew-Hamiltonian and irreducible.

[email protected] http://pages.pomona.edu/~sg064747

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Irreducible skew-Hamiltonians exist if n ≥ 8

Numerical Evidence

Computation indicates that generic 8× 8 SHMs are irreducible.Finding a scalable family of provably irreducible SHMs is difficult.

Proposition

For d ≥ 4, the matrix T =

(A B0 A

), where

A =

1

2. . .

d

, B =

0 1 1 · · · 1 1

−1 0 1. . . 1 1

−1 −1 0. . . 1 1

.... . .

. . .. . .

. . ....

−1 −1 −1. . . 0 1

−1 −1 −1 · · · −1 0

,

is skew-Hamiltonian and irreducible.

[email protected] http://pages.pomona.edu/~sg064747

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Putting It All

Together

[email protected] http://pages.pomona.edu/~sg064747

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An Instructive Example

A Potential Pitfall

T =

0 1 00 0 20 0 0

0 0 01 0 00 2 0

is UET, UECSM, and UESHM.

However, A =(

0 1 00 0 20 0 0

)itself has

none of these properties.

Remarks

If T = A⊕ At , then T is UECSM, UESHM, and UET.

Neither class UECSM nor UESHM is closed under restrictionto direct summands.

This issue is important only in dimensions ≥ 6

[email protected] http://pages.pomona.edu/~sg064747

Page 31: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

An Instructive Example

A Potential Pitfall

T =

0 1 00 0 20 0 0

0 0 01 0 00 2 0

is UET, UECSM, and UESHM. However, A =

(0 1 00 0 20 0 0

)itself has

none of these properties.

Remarks

If T = A⊕ At , then T is UECSM, UESHM, and UET.

Neither class UECSM nor UESHM is closed under restrictionto direct summands.

This issue is important only in dimensions ≥ 6

[email protected] http://pages.pomona.edu/~sg064747

Page 32: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

An Instructive Example

A Potential Pitfall

T =

0 1 00 0 20 0 0

0 0 01 0 00 2 0

is UET, UECSM, and UESHM. However, A =

(0 1 00 0 20 0 0

)itself has

none of these properties.

Remarks

If T = A⊕ At , then T is UECSM, UESHM, and UET.

Neither class UECSM nor UESHM is closed under restrictionto direct summands.

This issue is important only in dimensions ≥ 6

[email protected] http://pages.pomona.edu/~sg064747

Page 33: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

An Instructive Example

A Potential Pitfall

T =

0 1 00 0 20 0 0

0 0 01 0 00 2 0

is UET, UECSM, and UESHM. However, A =

(0 1 00 0 20 0 0

)itself has

none of these properties.

Remarks

If T = A⊕ At , then T is UECSM, UESHM, and UET.

Neither class UECSM nor UESHM is closed under restrictionto direct summands.

This issue is important only in dimensions ≥ 6

[email protected] http://pages.pomona.edu/~sg064747

Page 34: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

An Instructive Example

A Potential Pitfall

T =

0 1 00 0 20 0 0

0 0 01 0 00 2 0

is UET, UECSM, and UESHM. However, A =

(0 1 00 0 20 0 0

)itself has

none of these properties.

Remarks

If T = A⊕ At , then T is UECSM, UESHM, and UET.

Neither class UECSM nor UESHM is closed under restrictionto direct summands.

This issue is important only in dimensions ≥ 6

[email protected] http://pages.pomona.edu/~sg064747

Page 35: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

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Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

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Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 38: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 39: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 40: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 41: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM.

If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 42: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Final Answer

Theorem (SRG, JET)

T is UET if and only if T is unitarily equivalent to a direct sum of:

1 irreducible CSMs

2 irreducible SHMs (automatically 8× 8 or larger),

3 matrices of the form A⊕ At where A is irreducible and neitherUECSM nor UESHM (automatically 6× 6 or larger).

Moreover, the unitary orbits of these classes are pairwise disjoint.

Corollary

If T ∈ Mn(C) is UET and n ≤ 7, then T is UECSM. If n ≤ 5, thenT is unitarily equivalent to a direct sum of irreducible CSMs.

[email protected] http://pages.pomona.edu/~sg064747

Page 43: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A Trace Criterion

Theorem (Pearcy, 1962 + Sibirskiı, 1968)

For 3× 3 matrices, A ∼= B if and only if the traces of

X , X 2, X 3, X ∗X , X ∗X 2, X ∗2X 2, X ∗X 2X ∗2X

agree for X = A and X = B.

Proposition (SRG, JET)

If T is 3× 3, then T is UECSM if and only if

tr[T ∗T (T ∗T − TT ∗)TT ∗] = 0.

Proof.

Check if T ∼= T t . Only the final word is not satisfied trivially.

[email protected] http://pages.pomona.edu/~sg064747

Page 44: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A Trace Criterion

Theorem (Pearcy, 1962 + Sibirskiı, 1968)

For 3× 3 matrices, A ∼= B if and only if the traces of

X , X 2, X 3, X ∗X , X ∗X 2, X ∗2X 2, X ∗X 2X ∗2X

agree for X = A and X = B.

Proposition (SRG, JET)

If T is 3× 3, then T is UECSM if and only if

tr[T ∗T (T ∗T − TT ∗)TT ∗] = 0.

Proof.

Check if T ∼= T t . Only the final word is not satisfied trivially.

[email protected] http://pages.pomona.edu/~sg064747

Page 45: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

A Trace Criterion

Theorem (Pearcy, 1962 + Sibirskiı, 1968)

For 3× 3 matrices, A ∼= B if and only if the traces of

X , X 2, X 3, X ∗X , X ∗X 2, X ∗2X 2, X ∗X 2X ∗2X

agree for X = A and X = B.

Proposition (SRG, JET)

If T is 3× 3, then T is UECSM if and only if

tr[T ∗T (T ∗T − TT ∗)TT ∗] = 0.

Proof.

Check if T ∼= T t . Only the final word is not satisfied trivially.

[email protected] http://pages.pomona.edu/~sg064747

Page 46: On a problem of Halmos: Unitary equivalence of a matrix to ... · On a problem of Halmos: Unitary equivalence of a matrix to its transpose Stephan Ramon Garcia Pomona College Claremont,

Bibliography

1 Balayan, L., Garcia, S.R., Unitary equivalence to a complex symmetric matrix:geometric criteria, Oper. Matrices 4 (2010), No. 1, 53–76.

2 Garcia, S.R., Poore, D.E., Wyse, M.K., Unitary equivalent to a complex symmetricmatrix: a modulus criterion, Oper. Matrices (to appear).

3 Garcia, S.R., Tener, J.E., Unitary equivalence of a matrix to its transpose, J. OperatorTheory (to appear).

4 Garcia, S.R., Wogen, W., Complex symmetric partial isometries, J. Funct. Analysis257 (2009), 1251-1260.

5 Garcia, S.R., Wogen, W., Some new classes of complex symmetric operators, Trans.Amer. Math. Soc. 362 (2010), 6065-6077.

6 Halmos, P.R., Linear algebra problem book, The Dolciani Mathematical Expositions, 16.Mathematical Association of America, Washington, DC, 1995.

7 Pearcy, C., A complete set of unitary invariants for 3× 3 complex matrices, Trans. Amer.Math. Soc. 104 (1962), 425429.

8 Sibirskiı, K.S., A minimal polynomial basis of unitary invariants of a square matrix oforder three (Russian), Mat. Zametki 3 (1968), 291295.

9 Tener, J.E., Unitary equivalence to a complex symmetric matrix: an algorithm, J. Math.Anal. Appl. 341 (2008), 640–648.

10 Waterhouse, William C., The structure of alternating-Hamiltonian matrices, LinearAlgebra Appl. 396 (2005), 385390.

[email protected] http://pages.pomona.edu/~sg064747


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