Maximizing the trace of product of matrices under fixed spectrum
0
Is it correct that under fixed spectrum, $operatorname{tr}(AB)$ is maximized when $A$ and $B$ share the same eigenbasis? If yes, how can this be shown?
linear-algebra eigenvalues-eigenvectors maxima-minima trace
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
|
show 1 more comment
0
Is it correct that under fixed spectrum, $operatorname{tr}(AB)$ is maximized when $A$ and $B$ share the same eigenbasis? If yes, how can this be shown?
linear-algebra eigenvalues-eigenvectors maxima-minima trace
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46
|
show 1 more comment
0
0
0
Is it correct that under fixed spectrum, $operatorname{tr}(AB)$ is maximized when $A$ and $B$ share the same eigenbasis? If yes, how can this be shown?
linear-algebra eigenvalues-eigenvectors maxima-minima trace
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
Is it correct that under fixed spectrum, $operatorname{tr}(AB)$ is maximized when $A$ and $B$ share the same eigenbasis? If yes, how can this be shown?
linear-algebra eigenvalues-eigenvectors maxima-minima trace
linear-algebra eigenvalues-eigenvectors maxima-minima trace
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
edited Nov 26 at 22:31
Davide Giraudo
125k16150259
125k16150259
asked Nov 18 at 15:45
Desh Raj
11
asked Nov 18 at 15:45
Desh Raj
11
asked Nov 18 at 15:45
Desh Raj
11
11
Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46
|
show 1 more comment
Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46
Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46
|
show 1 more comment
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Yes, it is correct. The fact you are looking for is called the Von Neumann's trace inequality. I am not familiar with its proof(s).
– AnonymousCoward
Nov 26 at 23:02
@AnonymousCoward von Neumann's trace inequality is about singular values, not eigenvalues.
– user1551
Nov 27 at 4:39
"Under fixed spectrum" of what? Do you mean the spectrum of $AB$ is fixed? Or the spectra of $A$ and $B$ are fixed? Or something else?
– user1551
Nov 27 at 4:40
@AnonymousCoward Sorry, but I don't follow. Would you please be more specific?
– user1551
Nov 27 at 13:56
@user1551By fixed spectrum I meant that we fix the singular values of either matrix, say A.
– Desh Raj
Nov 27 at 23:46