Mixing
What are the conditions on $text{tr}(AB) leq text{tr(A)} text{tr(B)}$ to be true?
$begingroup$
Let $A$ and $B$ be two arbitrary matrix with proper dimension for multiplication.
Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces
$$text{tr}(AB) leq text{tr(A)} text{tr(B)}$$
1- Do we have result for rectangular matrix that satisfy this inequality?
2- If they were square matrices what are the conditions?
3- Is there any specific name for this inequality?
linear-algebra matrices trace
asked Jan 4 at 17:29
SaeedSaeed
969310
$endgroup$
add a comment |
$begingroup$
Let $A$ and $B$ be two arbitrary matrix with proper dimension for multiplication.
Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces
$$text{tr}(AB) leq text{tr(A)} text{tr(B)}$$
1- Do we have result for rectangular matrix that satisfy this inequality?
2- If they were square matrices what are the conditions?
3- Is there any specific name for this inequality?
linear-algebra matrices trace
asked Jan 4 at 17:29
SaeedSaeed
969310
$endgroup$
2
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46
add a comment |
$begingroup$
Let $A$ and $B$ be two arbitrary matrix with proper dimension for multiplication.
Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces
$$text{tr}(AB) leq text{tr(A)} text{tr(B)}$$
1- Do we have result for rectangular matrix that satisfy this inequality?
2- If they were square matrices what are the conditions?
3- Is there any specific name for this inequality?
linear-algebra matrices trace
asked Jan 4 at 17:29
SaeedSaeed
969310
$endgroup$
Let $A$ and $B$ be two arbitrary matrix with proper dimension for multiplication.
Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces
$$text{tr}(AB) leq text{tr(A)} text{tr(B)}$$
1- Do we have result for rectangular matrix that satisfy this inequality?
2- If they were square matrices what are the conditions?
3- Is there any specific name for this inequality?
linear-algebra matrices trace
linear-algebra matrices trace
asked Jan 4 at 17:29
SaeedSaeed
969310
asked Jan 4 at 17:29
SaeedSaeed
969310
asked Jan 4 at 17:29
SaeedSaeed
969310
asked Jan 4 at 17:29
SaeedSaeed
969310
asked Jan 4 at 17:29
SaeedSaeed
969310
969310
2
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46
add a comment |
2
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46
2
2
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true because
$$
operatorname{tr}(AB)
=operatorname{tr}(A^{1/2}A^{1/2}B)
=operatorname{tr}(A^{1/2}BA^{1/2})
=|A^{1/2}B^{1/2}|_F^2le|A^{1/2}|_F^2|B^{1/2}|_F^2
=operatorname{tr}(A)operatorname{tr}(B).
$$
answered Jan 4 at 17:43
user1551user1551
72.3k566127
$endgroup$
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
|
show 1 more comment
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1 Answer
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$begingroup$
Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true because
$$
operatorname{tr}(AB)
=operatorname{tr}(A^{1/2}A^{1/2}B)
=operatorname{tr}(A^{1/2}BA^{1/2})
=|A^{1/2}B^{1/2}|_F^2le|A^{1/2}|_F^2|B^{1/2}|_F^2
=operatorname{tr}(A)operatorname{tr}(B).
$$
answered Jan 4 at 17:43
user1551user1551
72.3k566127
$endgroup$
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
|
show 1 more comment
$begingroup$
Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true because
$$
operatorname{tr}(AB)
=operatorname{tr}(A^{1/2}A^{1/2}B)
=operatorname{tr}(A^{1/2}BA^{1/2})
=|A^{1/2}B^{1/2}|_F^2le|A^{1/2}|_F^2|B^{1/2}|_F^2
=operatorname{tr}(A)operatorname{tr}(B).
$$
answered Jan 4 at 17:43
user1551user1551
72.3k566127
$endgroup$
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
|
show 1 more comment
$begingroup$
Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true because
$$
operatorname{tr}(AB)
=operatorname{tr}(A^{1/2}A^{1/2}B)
=operatorname{tr}(A^{1/2}BA^{1/2})
=|A^{1/2}B^{1/2}|_F^2le|A^{1/2}|_F^2|B^{1/2}|_F^2
=operatorname{tr}(A)operatorname{tr}(B).
$$
answered Jan 4 at 17:43
user1551user1551
72.3k566127
$endgroup$
Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true because
$$
operatorname{tr}(AB)
=operatorname{tr}(A^{1/2}A^{1/2}B)
=operatorname{tr}(A^{1/2}BA^{1/2})
=|A^{1/2}B^{1/2}|_F^2le|A^{1/2}|_F^2|B^{1/2}|_F^2
=operatorname{tr}(A)operatorname{tr}(B).
$$
answered Jan 4 at 17:43
user1551user1551
72.3k566127
edited Jan 4 at 18:01
answered Jan 4 at 17:43
user1551user1551
72.3k566127
answered Jan 4 at 17:43
user1551user1551
72.3k566127
answered Jan 4 at 17:43
user1551user1551
72.3k566127
72.3k566127
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
|
show 1 more comment
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
is there any name for this inequality?
$endgroup$
– Saeed
Jan 4 at 17:51
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_ALambda_AU_A^T$ and $B=U_BLambda_BU_B^T$. So $text{tr}(U_ALambda_AU_A^TU_BLambda_BU_B^T)=text{tr}(U_B^TU_ALambda_AU_A^TU_BLambda_B)$, how can I get what you have?
$endgroup$
– Saeed
Jan 4 at 17:54
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
What's that $F$ norm?
$endgroup$
– Yanko
Jan 4 at 17:57
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Yanko $|cdot|_F$ denotes Frobenius norm.
$endgroup$
– user1551
Jan 4 at 18:00
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
$begingroup$
@Saeed $A^{1/2}$ is the (unique) positive semidefinite square root of $A$. If you orthogonally diagonalise $A$ as $QDQ^T$, then $A^{1/2}=QD^{1/2}Q^T$ where $D^{1/2}$ is the entrywise square root of $D$ (i.e. $D^{1/2}$ is the diagonal matrix whose diagonal entries are the square roots of their correspondents in $D$).
$endgroup$
– user1551
Jan 4 at 18:03
|
show 1 more comment
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2
$begingroup$
How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$sum_{i=1}^n lambda_i^2 leq left(sum_{i=1}^n lambda_i right)^2$$
$endgroup$
– Severin Schraven
Jan 4 at 17:39
$begingroup$
If we have a $mtimes n$-matrix, and $k:=min(m,n)$, the definition $$tr(A)=sum_{j=1}^k a_{jj}$$ would make sense.
$endgroup$
– Peter
Jan 4 at 17:43
$begingroup$
@Peter You are right, I was thinking about something which is coordinate-free.
$endgroup$
– Severin Schraven
Jan 4 at 17:45
$begingroup$
@Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3.
$endgroup$
– Saeed
Jan 4 at 17:46