Is $sqrtABsqrtA$ operator concave with respect to $A$?

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I want to show that $sqrtABsqrtA$ is operator concave with respect to $A$, when both $A$ and $B$ are Hermitian positive definite matrices.

I tested it numerically with different positive definite matrices, using the following condition:
$$sqrtfracA+C2BsqrtfracA+C2>fracsqrtCBsqrtC2+fracsqrtABsqrtA2$$

and it seems that it is true, but I don't know how to prove it.

Update: from an answer below this post, it turns out that $sqrtABsqrtA$ is NOT operator concave.

linear-algebra matrices

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edited Jul 17 at 21:31

asked Jul 16 at 4:10

Mah

550312

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up vote
3
down vote

favorite

I want to show that $sqrtABsqrtA$ is operator concave with respect to $A$, when both $A$ and $B$ are Hermitian positive definite matrices.

I tested it numerically with different positive definite matrices, using the following condition:
$$sqrtfracA+C2BsqrtfracA+C2>fracsqrtCBsqrtC2+fracsqrtABsqrtA2$$

and it seems that it is true, but I don't know how to prove it.

Update: from an answer below this post, it turns out that $sqrtABsqrtA$ is NOT operator concave.

linear-algebra matrices

share|cite|improve this question

edited Jul 17 at 21:31

asked Jul 16 at 4:10

Mah

550312

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I want to show that $sqrtABsqrtA$ is operator concave with respect to $A$, when both $A$ and $B$ are Hermitian positive definite matrices.

I tested it numerically with different positive definite matrices, using the following condition:
$$sqrtfracA+C2BsqrtfracA+C2>fracsqrtCBsqrtC2+fracsqrtABsqrtA2$$

and it seems that it is true, but I don't know how to prove it.

Update: from an answer below this post, it turns out that $sqrtABsqrtA$ is NOT operator concave.

linear-algebra matrices

share|cite|improve this question

edited Jul 17 at 21:31

asked Jul 16 at 4:10

Mah

550312

I want to show that $sqrtABsqrtA$ is operator concave with respect to $A$, when both $A$ and $B$ are Hermitian positive definite matrices.

I tested it numerically with different positive definite matrices, using the following condition:
$$sqrtfracA+C2BsqrtfracA+C2>fracsqrtCBsqrtC2+fracsqrtABsqrtA2$$

and it seems that it is true, but I don't know how to prove it.

Update: from an answer below this post, it turns out that $sqrtABsqrtA$ is NOT operator concave.

linear-algebra matrices

share|cite|improve this question

edited Jul 17 at 21:31

asked Jul 16 at 4:10

Mah

550312

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 21:31

edited Jul 17 at 21:31

edited Jul 17 at 21:31

asked Jul 16 at 4:10

Mah

550312

asked Jul 16 at 4:10

Mah

550312

asked Jul 16 at 4:10

Mah

550312

550312

add a comment | 

add a comment | 

1 Answer
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2
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accepted

The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=operatornamediag(2,0),C=operatornamediag(0,2)$ and $B=pmatrix2&-1\ -1&2$. The inequality then becomes
$$
pmatrix2&-1\ -1&2gepmatrix0&0\ 0&2+pmatrix2&0\ 0&0=2I,
$$
which is not true.

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=operatornamediag(2,0),C=operatornamediag(0,2)$ and $B=pmatrix2&-1\ -1&2$. The inequality then becomes
$$
pmatrix2&-1\ -1&2gepmatrix0&0\ 0&2+pmatrix2&0\ 0&0=2I,
$$
which is not true.

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

66.9k564122

add a comment | 

up vote
2
down vote



accepted

The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=operatornamediag(2,0),C=operatornamediag(0,2)$ and $B=pmatrix2&-1\ -1&2$. The inequality then becomes
$$
pmatrix2&-1\ -1&2gepmatrix0&0\ 0&2+pmatrix2&0\ 0&0=2I,
$$
which is not true.

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

66.9k564122

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=operatornamediag(2,0),C=operatornamediag(0,2)$ and $B=pmatrix2&-1\ -1&2$. The inequality then becomes
$$
pmatrix2&-1\ -1&2gepmatrix0&0\ 0&2+pmatrix2&0\ 0&0=2I,
$$
which is not true.

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

66.9k564122

The inequality does not always hold. If it holds in general, by passing $A$ and $C$ to the limit, it should hold (with $ge$ in place of $>$) for every pair of positive semidefinite matrices $A,C$ and every positive definite $B$. Let $A=operatornamediag(2,0),C=operatornamediag(0,2)$ and $B=pmatrix2&-1\ -1&2$. The inequality then becomes
$$
pmatrix2&-1\ -1&2gepmatrix0&0\ 0&2+pmatrix2&0\ 0&0=2I,
$$
which is not true.

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

66.9k564122

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 5:26

user1551

66.9k564122

answered Jul 16 at 5:26

user1551

66.9k564122

answered Jul 16 at 5:26

user1551

66.9k564122

66.9k564122

add a comment | 

add a comment | 

 

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