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Commutativity of trace norm

How to prove $operatornameTr(AB) = operatornameTr(BA)$?Trace class for operatorsAbout the trace class operators and their motivationThose differential operators that are bounded.Properties of trace-class operatorsQuestion about norm in trace classTrace class norm and rank inequalityHolders inequality for Hilbert Schmidt operators which are also trace classPartial trace for nuclear operators on a projective tensor productAbsolute convergence of trace of trace class operatorTrace norm inequality involving partial trace

1

$begingroup$

Please, does anyone knows whether $operatornametr |AB| = operatornametr |BA|$ for general trace operators $A$ and $B$ in Hilbert spaces ? Do you have a counter-example or a proof ?

Thanks

ps : I know $operatornametr AB = operatornametr BA$

operator-theory

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edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

$endgroup$

add a comment | 

1

$begingroup$

Please, does anyone knows whether $operatornametr |AB| = operatornametr |BA|$ for general trace operators $A$ and $B$ in Hilbert spaces ? Do you have a counter-example or a proof ?

Thanks

ps : I know $operatornametr AB = operatornametr BA$

operator-theory

share|cite|improve this question

edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

$endgroup$

add a comment | 

$begingroup$

Please, does anyone knows whether $operatornametr |AB| = operatornametr |BA|$ for general trace operators $A$ and $B$ in Hilbert spaces ? Do you have a counter-example or a proof ?

Thanks

ps : I know $operatornametr AB = operatornametr BA$

operator-theory

share|cite|improve this question

edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

share|cite|improve this question

edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

edited Mar 11 at 20:02

Bernard

123k741116

edited Mar 11 at 20:02

Bernard

123k741116

asked Mar 11 at 19:47

LouiseLouise

63

asked Mar 11 at 19:47

LouiseLouise

63

2 Answers
2

active

oldest

votes

2

$begingroup$

Consider the (randomly chosen) $2 times 2$ matrices
$$ A = pmatrix0 & 2cr 2 & 3cr, B = pmatrix1 & 0cr -3 & -2$$
I get
$$|AB| = sqrt(AB)^* AB = frac1sqrt17 pmatrix31 & 22cr 22 & 20$$
$$ |BA| = sqrt(BA)^* BA = frac1sqrt5 pmatrix4 & 8cr 8 & 26$$
with traces $3sqrt17$ and $6 sqrt5$ which are different.

share|cite|improve this answer

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

$endgroup$

add a comment | 

0

$begingroup$

For a very easy example, take $$A=beginbmatrix 1&0\0&0endbmatrix, B=beginbmatrix 0&1\0&0endbmatrix.$$
Then
$$
AB=B, BA=0.
$$
Thus, as $|AB|=|B|=I-A$, we have
$$0=operatornameTr(|BA|)ne1=operatornameTr(|AB|).$$

share|cite|improve this answer

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

add a comment | 

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2

$begingroup$

Consider the (randomly chosen) $2 times 2$ matrices
$$ A = pmatrix0 & 2cr 2 & 3cr, B = pmatrix1 & 0cr -3 & -2$$
I get
$$|AB| = sqrt(AB)^* AB = frac1sqrt17 pmatrix31 & 22cr 22 & 20$$
$$ |BA| = sqrt(BA)^* BA = frac1sqrt5 pmatrix4 & 8cr 8 & 26$$
with traces $3sqrt17$ and $6 sqrt5$ which are different.

share|cite|improve this answer

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

$endgroup$

add a comment | 

2

$begingroup$

Consider the (randomly chosen) $2 times 2$ matrices
$$ A = pmatrix0 & 2cr 2 & 3cr, B = pmatrix1 & 0cr -3 & -2$$
I get
$$|AB| = sqrt(AB)^* AB = frac1sqrt17 pmatrix31 & 22cr 22 & 20$$
$$ |BA| = sqrt(BA)^* BA = frac1sqrt5 pmatrix4 & 8cr 8 & 26$$
with traces $3sqrt17$ and $6 sqrt5$ which are different.

share|cite|improve this answer

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

$endgroup$

add a comment | 

$begingroup$

Consider the (randomly chosen) $2 times 2$ matrices
$$ A = pmatrix0 & 2cr 2 & 3cr, B = pmatrix1 & 0cr -3 & -2$$
I get
$$|AB| = sqrt(AB)^* AB = frac1sqrt17 pmatrix31 & 22cr 22 & 20$$
$$ |BA| = sqrt(BA)^* BA = frac1sqrt5 pmatrix4 & 8cr 8 & 26$$
with traces $3sqrt17$ and $6 sqrt5$ which are different.

share|cite|improve this answer

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

answered Mar 11 at 20:29

Robert IsraelRobert Israel

327k23216470

0

$begingroup$

For a very easy example, take $$A=beginbmatrix 1&0\0&0endbmatrix, B=beginbmatrix 0&1\0&0endbmatrix.$$
Then
$$
AB=B, BA=0.
$$
Thus, as $|AB|=|B|=I-A$, we have
$$0=operatornameTr(|BA|)ne1=operatornameTr(|AB|).$$

share|cite|improve this answer

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

add a comment | 

0

$begingroup$

For a very easy example, take $$A=beginbmatrix 1&0\0&0endbmatrix, B=beginbmatrix 0&1\0&0endbmatrix.$$
Then
$$
AB=B, BA=0.
$$
Thus, as $|AB|=|B|=I-A$, we have
$$0=operatornameTr(|BA|)ne1=operatornameTr(|AB|).$$

share|cite|improve this answer

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

add a comment | 

$begingroup$

For a very easy example, take $$A=beginbmatrix 1&0\0&0endbmatrix, B=beginbmatrix 0&1\0&0endbmatrix.$$
Then
$$
AB=B, BA=0.
$$
Thus, as $|AB|=|B|=I-A$, we have
$$0=operatornameTr(|BA|)ne1=operatornameTr(|AB|).$$

share|cite|improve this answer

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

$endgroup$

share|cite|improve this answer

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184

answered Mar 12 at 21:19

Martin ArgeramiMartin Argerami

128k1184184