How to prove this matrix inequality？

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Given the following two conditions:

(1). $A$ and $B$ are $ntimes n$ matrices;

(2). $U$ is a unitary $ntimes n$ matrix, i.e., $UU^+=U^+U=1$

How to prove this matrix inequality:
tr$left( AUBU^+right) leq $tr$left( ABright) $ ?

matrices inequality

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asked Jul 18 at 22:35

Daniel Kim

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0
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favorite

Given the following two conditions:

(1). $A$ and $B$ are $ntimes n$ matrices;

(2). $U$ is a unitary $ntimes n$ matrix, i.e., $UU^+=U^+U=1$

How to prove this matrix inequality:
tr$left( AUBU^+right) leq $tr$left( ABright) $ ?

matrices inequality

share|cite|improve this question

asked Jul 18 at 22:35

Daniel Kim

213

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Given the following two conditions:

(1). $A$ and $B$ are $ntimes n$ matrices;

(2). $U$ is a unitary $ntimes n$ matrix, i.e., $UU^+=U^+U=1$

How to prove this matrix inequality:
tr$left( AUBU^+right) leq $tr$left( ABright) $ ?

matrices inequality

share|cite|improve this question

asked Jul 18 at 22:35

Daniel Kim

213

Given the following two conditions:

(1). $A$ and $B$ are $ntimes n$ matrices;

(2). $U$ is a unitary $ntimes n$ matrix, i.e., $UU^+=U^+U=1$

How to prove this matrix inequality:
tr$left( AUBU^+right) leq $tr$left( ABright) $ ?

matrices inequality

share|cite|improve this question

asked Jul 18 at 22:35

Daniel Kim

213

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 22:35

Daniel Kim

213

asked Jul 18 at 22:35

Daniel Kim

213

asked Jul 18 at 22:35

Daniel Kim

213

213

add a comment | 

add a comment | 

2 Answers
2

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1
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Here is a proof for real matrices. I don't think it makes sense to prove this for complex matrices as trace can then be complex valued and they cannot be ordered.

By Spectral Theorem for normal matrices, we get that for unitary matrix $U$ is diagonalizable and in particular all of its eigenvalues have radius 1. For the proof, observe the following: let $U$ be unitary and $Uv=lambda v$. Then,

$$||Uv||^2 = langle Uv, Uv rangle = langle v, U^*Uvrangle= ||v||^2$$
Also,
$$||Uv||^2 = |lambda|^2 ||v||^2$$

Hence,$ |lambda| = 1$ or $lambda=-1,1$.

Then, using properties of trace,

$$tr(AUBU^*) leq tr(AUB)tr(U^*) = tr(UBA)tr(U^*) leq tr(U)tr(U^*)tr(BA) = tr(AB)$$

where the last equality is from the observation that trace of a matrix is the sum of eigenvalues and that for unitary matrix $U$ and $U^*$, the eigenvalues are again $-1,1$ with same multiplicities.

EDIT:
The last assertion still needs work and may need more assumptions.

share|cite|improve this answer

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Trace of an identity is not $1$ except for a very special case.
  – Algebraic Pavel
  Jul 18 at 23:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I realized that moments after I answered.
  – James Yang
  Jul 18 at 23:34
  

  
  
  
  

add a comment | 

up vote
0
down vote

This is only true for Hermitian matrices and does not need to hold otherwise.
For example, if
$$
A=B=beginbmatrix1 & 1\0 & 1endbmatrix,
quad
U=beginbmatrix0 & 1\ 1 & 0endbmatrix,
$$
then
$$
mathrmtrace(AUBU^*)=3notleq 2=mathrmtrace(AB).
$$

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Here is a proof for real matrices. I don't think it makes sense to prove this for complex matrices as trace can then be complex valued and they cannot be ordered.

By Spectral Theorem for normal matrices, we get that for unitary matrix $U$ is diagonalizable and in particular all of its eigenvalues have radius 1. For the proof, observe the following: let $U$ be unitary and $Uv=lambda v$. Then,

$$||Uv||^2 = langle Uv, Uv rangle = langle v, U^*Uvrangle= ||v||^2$$
Also,
$$||Uv||^2 = |lambda|^2 ||v||^2$$

Hence,$ |lambda| = 1$ or $lambda=-1,1$.

Then, using properties of trace,

$$tr(AUBU^*) leq tr(AUB)tr(U^*) = tr(UBA)tr(U^*) leq tr(U)tr(U^*)tr(BA) = tr(AB)$$

where the last equality is from the observation that trace of a matrix is the sum of eigenvalues and that for unitary matrix $U$ and $U^*$, the eigenvalues are again $-1,1$ with same multiplicities.

EDIT:
The last assertion still needs work and may need more assumptions.

share|cite|improve this answer

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Trace of an identity is not $1$ except for a very special case.
  – Algebraic Pavel
  Jul 18 at 23:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I realized that moments after I answered.
  – James Yang
  Jul 18 at 23:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

Here is a proof for real matrices. I don't think it makes sense to prove this for complex matrices as trace can then be complex valued and they cannot be ordered.

By Spectral Theorem for normal matrices, we get that for unitary matrix $U$ is diagonalizable and in particular all of its eigenvalues have radius 1. For the proof, observe the following: let $U$ be unitary and $Uv=lambda v$. Then,

$$||Uv||^2 = langle Uv, Uv rangle = langle v, U^*Uvrangle= ||v||^2$$
Also,
$$||Uv||^2 = |lambda|^2 ||v||^2$$

Hence,$ |lambda| = 1$ or $lambda=-1,1$.

Then, using properties of trace,

$$tr(AUBU^*) leq tr(AUB)tr(U^*) = tr(UBA)tr(U^*) leq tr(U)tr(U^*)tr(BA) = tr(AB)$$

where the last equality is from the observation that trace of a matrix is the sum of eigenvalues and that for unitary matrix $U$ and $U^*$, the eigenvalues are again $-1,1$ with same multiplicities.

EDIT:
The last assertion still needs work and may need more assumptions.

share|cite|improve this answer

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Trace of an identity is not $1$ except for a very special case.
  – Algebraic Pavel
  Jul 18 at 23:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I realized that moments after I answered.
  – James Yang
  Jul 18 at 23:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Here is a proof for real matrices. I don't think it makes sense to prove this for complex matrices as trace can then be complex valued and they cannot be ordered.

By Spectral Theorem for normal matrices, we get that for unitary matrix $U$ is diagonalizable and in particular all of its eigenvalues have radius 1. For the proof, observe the following: let $U$ be unitary and $Uv=lambda v$. Then,

$$||Uv||^2 = langle Uv, Uv rangle = langle v, U^*Uvrangle= ||v||^2$$
Also,
$$||Uv||^2 = |lambda|^2 ||v||^2$$

Hence,$ |lambda| = 1$ or $lambda=-1,1$.

Then, using properties of trace,

$$tr(AUBU^*) leq tr(AUB)tr(U^*) = tr(UBA)tr(U^*) leq tr(U)tr(U^*)tr(BA) = tr(AB)$$

where the last equality is from the observation that trace of a matrix is the sum of eigenvalues and that for unitary matrix $U$ and $U^*$, the eigenvalues are again $-1,1$ with same multiplicities.

EDIT:
The last assertion still needs work and may need more assumptions.

share|cite|improve this answer

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

Here is a proof for real matrices. I don't think it makes sense to prove this for complex matrices as trace can then be complex valued and they cannot be ordered.

By Spectral Theorem for normal matrices, we get that for unitary matrix $U$ is diagonalizable and in particular all of its eigenvalues have radius 1. For the proof, observe the following: let $U$ be unitary and $Uv=lambda v$. Then,

$$||Uv||^2 = langle Uv, Uv rangle = langle v, U^*Uvrangle= ||v||^2$$
Also,
$$||Uv||^2 = |lambda|^2 ||v||^2$$

Hence,$ |lambda| = 1$ or $lambda=-1,1$.

Then, using properties of trace,

$$tr(AUBU^*) leq tr(AUB)tr(U^*) = tr(UBA)tr(U^*) leq tr(U)tr(U^*)tr(BA) = tr(AB)$$

where the last equality is from the observation that trace of a matrix is the sum of eigenvalues and that for unitary matrix $U$ and $U^*$, the eigenvalues are again $-1,1$ with same multiplicities.

EDIT:
The last assertion still needs work and may need more assumptions.

share|cite|improve this answer

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 23:26

edited Jul 18 at 23:26

edited Jul 18 at 23:26

answered Jul 18 at 23:08

James Yang

4349

answered Jul 18 at 23:08

James Yang

4349

answered Jul 18 at 23:08

James Yang

4349

4349

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Trace of an identity is not $1$ except for a very special case.
  – Algebraic Pavel
  Jul 18 at 23:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I realized that moments after I answered.
  – James Yang
  Jul 18 at 23:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Trace of an identity is not $1$ except for a very special case.
  – Algebraic Pavel
  Jul 18 at 23:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I realized that moments after I answered.
  – James Yang
  Jul 18 at 23:34
  

  
  
  
  

Trace of an identity is not $1$ except for a very special case.
– Algebraic Pavel
Jul 18 at 23:33

Trace of an identity is not $1$ except for a very special case.
– Algebraic Pavel
Jul 18 at 23:33

Yes I realized that moments after I answered.
– James Yang
Jul 18 at 23:34

Yes I realized that moments after I answered.
– James Yang
Jul 18 at 23:34

add a comment | 

up vote
0
down vote

This is only true for Hermitian matrices and does not need to hold otherwise.
For example, if
$$
A=B=beginbmatrix1 & 1\0 & 1endbmatrix,
quad
U=beginbmatrix0 & 1\ 1 & 0endbmatrix,
$$
then
$$
mathrmtrace(AUBU^*)=3notleq 2=mathrmtrace(AB).
$$

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

add a comment | 

up vote
0
down vote

This is only true for Hermitian matrices and does not need to hold otherwise.
For example, if
$$
A=B=beginbmatrix1 & 1\0 & 1endbmatrix,
quad
U=beginbmatrix0 & 1\ 1 & 0endbmatrix,
$$
then
$$
mathrmtrace(AUBU^*)=3notleq 2=mathrmtrace(AB).
$$

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

add a comment | 

up vote
0
down vote

up vote
0
down vote

This is only true for Hermitian matrices and does not need to hold otherwise.
For example, if
$$
A=B=beginbmatrix1 & 1\0 & 1endbmatrix,
quad
U=beginbmatrix0 & 1\ 1 & 0endbmatrix,
$$
then
$$
mathrmtrace(AUBU^*)=3notleq 2=mathrmtrace(AB).
$$

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

This is only true for Hermitian matrices and does not need to hold otherwise.
For example, if
$$
A=B=beginbmatrix1 & 1\0 & 1endbmatrix,
quad
U=beginbmatrix0 & 1\ 1 & 0endbmatrix,
$$
then
$$
mathrmtrace(AUBU^*)=3notleq 2=mathrmtrace(AB).
$$

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

answered Jul 20 at 9:07

Algebraic Pavel

15.7k31638

15.7k31638

add a comment | 

add a comment | 

 

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