operator concavity of a function involving trace and logarithm

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I want to see if
$$f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$$
is operator concave with respect to a Hermitian positive definite matrices $A$?
$log$ is matrix logarithm, $B$ and $C$ are arbitrary Hermitian positive definite matrices, and $I$ is unit matrix.

I checked it numerically with many randomly generated positive definite matrices, using the following condition from Bhatia's matrix analysis:
$$fleft(fracX+Y2right)>fracf(X)2+fracf(Y)2$$
and it seems that the condition is satisfied.

I want to prove it using the same condition. I am trying to show:
$$ operatornametraceleft(Clogleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)>fracoperatornametraceleft(Cleft(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)2, (1)$$

Update:

First, I solve it for special case $C=I$ (i.e. $C$ is identity matrix). Using $operatornametrace(log x)=log det(x)$, left side of (1) is,
$$operatornametraceleft(logleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+fracX+Y2Bright)right), (2)$$
Third line is from Sylvester's determinant identity(https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity). Right side of (1) becomes
$$frac12operatornametraceleft(left(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+sqrtXBsqrtX)right)+logleft(det(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+XB)right)+logleft(det(I+YB)right)right), (3)$$

From concavity of log-determinant (Log-Determinant Concavity Proof), I conclude (2) is greater than(3) and thus the condition (1) is satisfied for $C=I$. I have two questions:

1) Is my conclusion correct?(I am not 100% sure)

2)How can I extend this result (if it is correct) to a general positive definite $C$?

Update 2: I also tried the approach suggested in the answer to this question: Is the trace of inverse matrix convex? ...but the second derivative was too complicated.

Any help is very appreciated.

If you are aware of any other way to prove concavity of $f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$ please let me know.

linear-algebra matrices convex-analysis trace positive-definite

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edited Jul 20 at 3:08

asked Jul 17 at 21:49

Mah

545312

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  Don't you think you get more answers if you explain the notions you use?
  – amsmath
  Jul 17 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated my question and tried to explain things clearly.
  – Mah
  Jul 19 at 12:42
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

3

I want to see if
$$f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$$
is operator concave with respect to a Hermitian positive definite matrices $A$?
$log$ is matrix logarithm, $B$ and $C$ are arbitrary Hermitian positive definite matrices, and $I$ is unit matrix.

I checked it numerically with many randomly generated positive definite matrices, using the following condition from Bhatia's matrix analysis:
$$fleft(fracX+Y2right)>fracf(X)2+fracf(Y)2$$
and it seems that the condition is satisfied.

I want to prove it using the same condition. I am trying to show:
$$ operatornametraceleft(Clogleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)>fracoperatornametraceleft(Cleft(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)2, (1)$$

Update:

First, I solve it for special case $C=I$ (i.e. $C$ is identity matrix). Using $operatornametrace(log x)=log det(x)$, left side of (1) is,
$$operatornametraceleft(logleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+fracX+Y2Bright)right), (2)$$
Third line is from Sylvester's determinant identity(https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity). Right side of (1) becomes
$$frac12operatornametraceleft(left(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+sqrtXBsqrtX)right)+logleft(det(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+XB)right)+logleft(det(I+YB)right)right), (3)$$

From concavity of log-determinant (Log-Determinant Concavity Proof), I conclude (2) is greater than(3) and thus the condition (1) is satisfied for $C=I$. I have two questions:

1) Is my conclusion correct?(I am not 100% sure)

2)How can I extend this result (if it is correct) to a general positive definite $C$?

Update 2: I also tried the approach suggested in the answer to this question: Is the trace of inverse matrix convex? ...but the second derivative was too complicated.

Any help is very appreciated.

If you are aware of any other way to prove concavity of $f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$ please let me know.

linear-algebra matrices convex-analysis trace positive-definite

share|cite|improve this question

edited Jul 20 at 3:08

asked Jul 17 at 21:49

Mah

545312

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Don't you think you get more answers if you explain the notions you use?
  – amsmath
  Jul 17 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated my question and tried to explain things clearly.
  – Mah
  Jul 19 at 12:42
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

3

up vote
4
down vote

favorite

3

3

I want to see if
$$f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$$
is operator concave with respect to a Hermitian positive definite matrices $A$?
$log$ is matrix logarithm, $B$ and $C$ are arbitrary Hermitian positive definite matrices, and $I$ is unit matrix.

I checked it numerically with many randomly generated positive definite matrices, using the following condition from Bhatia's matrix analysis:
$$fleft(fracX+Y2right)>fracf(X)2+fracf(Y)2$$
and it seems that the condition is satisfied.

I want to prove it using the same condition. I am trying to show:
$$ operatornametraceleft(Clogleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)>fracoperatornametraceleft(Cleft(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)2, (1)$$

Update:

First, I solve it for special case $C=I$ (i.e. $C$ is identity matrix). Using $operatornametrace(log x)=log det(x)$, left side of (1) is,
$$operatornametraceleft(logleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+fracX+Y2Bright)right), (2)$$
Third line is from Sylvester's determinant identity(https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity). Right side of (1) becomes
$$frac12operatornametraceleft(left(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+sqrtXBsqrtX)right)+logleft(det(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+XB)right)+logleft(det(I+YB)right)right), (3)$$

From concavity of log-determinant (Log-Determinant Concavity Proof), I conclude (2) is greater than(3) and thus the condition (1) is satisfied for $C=I$. I have two questions:

1) Is my conclusion correct?(I am not 100% sure)

2)How can I extend this result (if it is correct) to a general positive definite $C$?

Update 2: I also tried the approach suggested in the answer to this question: Is the trace of inverse matrix convex? ...but the second derivative was too complicated.

Any help is very appreciated.

If you are aware of any other way to prove concavity of $f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$ please let me know.

linear-algebra matrices convex-analysis trace positive-definite

share|cite|improve this question

edited Jul 20 at 3:08

asked Jul 17 at 21:49

Mah

545312

I want to see if
$$f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$$
is operator concave with respect to a Hermitian positive definite matrices $A$?
$log$ is matrix logarithm, $B$ and $C$ are arbitrary Hermitian positive definite matrices, and $I$ is unit matrix.

I checked it numerically with many randomly generated positive definite matrices, using the following condition from Bhatia's matrix analysis:
$$fleft(fracX+Y2right)>fracf(X)2+fracf(Y)2$$
and it seems that the condition is satisfied.

I want to prove it using the same condition. I am trying to show:
$$ operatornametraceleft(Clogleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)>fracoperatornametraceleft(Cleft(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)2, (1)$$

Update:

First, I solve it for special case $C=I$ (i.e. $C$ is identity matrix). Using $operatornametrace(log x)=log det(x)$, left side of (1) is,
$$operatornametraceleft(logleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+sqrtfracX+Y2BsqrtfracX+Y2right)right)
\=logleft(detleft(I+fracX+Y2Bright)right), (2)$$
Third line is from Sylvester's determinant identity(https://en.wikipedia.org/wiki/Sylvester%27s_determinant_identity). Right side of (1) becomes
$$frac12operatornametraceleft(left(log(I+sqrtXBsqrtX)+log(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+sqrtXBsqrtX)right)+logleft(det(I+sqrtYBsqrtY)right)right)\
=frac12left(logleft(det(I+XB)right)+logleft(det(I+YB)right)right), (3)$$

From concavity of log-determinant (Log-Determinant Concavity Proof), I conclude (2) is greater than(3) and thus the condition (1) is satisfied for $C=I$. I have two questions:

1) Is my conclusion correct?(I am not 100% sure)

2)How can I extend this result (if it is correct) to a general positive definite $C$?

Update 2: I also tried the approach suggested in the answer to this question: Is the trace of inverse matrix convex? ...but the second derivative was too complicated.

Any help is very appreciated.

If you are aware of any other way to prove concavity of $f(A)= operatornametrace(Clog(I+sqrtABsqrtA))$ please let me know.

linear-algebra matrices convex-analysis trace positive-definite

share|cite|improve this question

edited Jul 20 at 3:08

asked Jul 17 at 21:49

Mah

545312

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 3:08

edited Jul 20 at 3:08

edited Jul 20 at 3:08

asked Jul 17 at 21:49

Mah

545312

asked Jul 17 at 21:49

Mah

545312

asked Jul 17 at 21:49

Mah

545312

545312

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Don't you think you get more answers if you explain the notions you use?
  – amsmath
  Jul 17 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated my question and tried to explain things clearly.
  – Mah
  Jul 19 at 12:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Don't you think you get more answers if you explain the notions you use?
  – amsmath
  Jul 17 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated my question and tried to explain things clearly.
  – Mah
  Jul 19 at 12:42
  

  
  
  
  

1

1

Don't you think you get more answers if you explain the notions you use?
– amsmath
Jul 17 at 21:59

Don't you think you get more answers if you explain the notions you use?
– amsmath
Jul 17 at 21:59

I updated my question and tried to explain things clearly.
– Mah
Jul 19 at 12:42

I updated my question and tried to explain things clearly.
– Mah
Jul 19 at 12:42

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

+50

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
 0.485971910201617 0.392980465973661
 0.392980465973661 0.323621544804127
>> A2
A2 =
 0.864345280133754 0.504105821991426
 0.504105821991426 0.596439049805159
>> B
B =
 1.659740432896511 0.467165733847868
 0.467165733847868 0.487112266549647
>> C
C =
 0.132518904109893 0.298931955806039
 0.298931955806039 1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

+50

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
 0.485971910201617 0.392980465973661
 0.392980465973661 0.323621544804127
>> A2
A2 =
 0.864345280133754 0.504105821991426
 0.504105821991426 0.596439049805159
>> B
B =
 1.659740432896511 0.467165733847868
 0.467165733847868 0.487112266549647
>> C
C =
 0.132518904109893 0.298931955806039
 0.298931955806039 1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

add a comment | 

up vote
0
down vote



accepted

+50

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
 0.485971910201617 0.392980465973661
 0.392980465973661 0.323621544804127
>> A2
A2 =
 0.864345280133754 0.504105821991426
 0.504105821991426 0.596439049805159
>> B
B =
 1.659740432896511 0.467165733847868
 0.467165733847868 0.487112266549647
>> C
C =
 0.132518904109893 0.298931955806039
 0.298931955806039 1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

add a comment | 

up vote
0
down vote



accepted

+50

up vote
0
down vote



accepted

+50

+50

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
 0.485971910201617 0.392980465973661
 0.392980465973661 0.323621544804127
>> A2
A2 =
 0.864345280133754 0.504105821991426
 0.504105821991426 0.596439049805159
>> B
B =
 1.659740432896511 0.467165733847868
 0.467165733847868 0.487112266549647
>> C
C =
 0.132518904109893 0.298931955806039
 0.298931955806039 1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

I also tried this numerically and quickly stumbled upon counterexamples

>> A1
A1 =
 0.485971910201617 0.392980465973661
 0.392980465973661 0.323621544804127
>> A2
A2 =
 0.864345280133754 0.504105821991426
 0.504105821991426 0.596439049805159
>> B
B =
 1.659740432896511 0.467165733847868
 0.467165733847868 0.487112266549647
>> C
C =
 0.132518904109893 0.298931955806039
 0.298931955806039 1.161589772124985
>> A3 = (A1+A2)/2;
>> f1 = trace(C*logm(I+sqrtm(A1)*B*sqrtm(A1)))
f1 = 0.6891
>> f2 = trace(C*logm(I+sqrtm(A2)*B*sqrtm(A2)))
f2 = 0.8090
>> f3 = trace(C*logm(I+sqrtm(A3)*B*sqrtm(A3)))
f3 = 0.7355

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 13:05

LinAlg

5,4061319

answered Jul 23 at 13:05

LinAlg

5,4061319

answered Jul 23 at 13:05

LinAlg

5,4061319

5,4061319

add a comment | 

add a comment | 

 

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