Mixing
On what condition $trace(A) ge trace(AB)$?
$begingroup$
Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) ge trace(AB)$?
linear-algebra matrix-calculus positive-semidefinite
asked Jan 20 at 17:46
BabakBabak
354111
$endgroup$
add a comment |
$begingroup$
Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) ge trace(AB)$?
linear-algebra matrix-calculus positive-semidefinite
asked Jan 20 at 17:46
BabakBabak
354111
$endgroup$
$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53
add a comment |
$begingroup$
Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) ge trace(AB)$?
linear-algebra matrix-calculus positive-semidefinite
asked Jan 20 at 17:46
BabakBabak
354111
$endgroup$
Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) ge trace(AB)$?
linear-algebra matrix-calculus positive-semidefinite
linear-algebra matrix-calculus positive-semidefinite
asked Jan 20 at 17:46
BabakBabak
354111
asked Jan 20 at 17:46
BabakBabak
354111
edited Jan 20 at 17:50
Babak
asked Jan 20 at 17:46
BabakBabak
354111
asked Jan 20 at 17:46
BabakBabak
354111
asked Jan 20 at 17:46
BabakBabak
354111
354111
$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53
add a comment |
$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53
$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $lambda_1,.., lambda_n$ are the eigenvalues of $A$ and $beta_1,.., beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $sigma$ such that
$$tr(AB)=lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $lambda_1,.., lambda_n$ and $beta_1,.., beta_n$, (and a permutation $sigma$ which can actually be ignored by reordering $beta$) under which condition do we have
$$lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}leq lambda_1+..+lambda_n ?$$
Note that for any such $lambda, beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $lambda_1,.., lambda_n$ are the eigenvalues of $A$ and $beta_1,.., beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $sigma$ such that
$$tr(AB)=lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $lambda_1,.., lambda_n$ and $beta_1,.., beta_n$, (and a permutation $sigma$ which can actually be ignored by reordering $beta$) under which condition do we have
$$lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}leq lambda_1+..+lambda_n ?$$
Note that for any such $lambda, beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
add a comment |
$begingroup$
Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $lambda_1,.., lambda_n$ are the eigenvalues of $A$ and $beta_1,.., beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $sigma$ such that
$$tr(AB)=lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $lambda_1,.., lambda_n$ and $beta_1,.., beta_n$, (and a permutation $sigma$ which can actually be ignored by reordering $beta$) under which condition do we have
$$lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}leq lambda_1+..+lambda_n ?$$
Note that for any such $lambda, beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
$endgroup$
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
add a comment |
$begingroup$
Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $lambda_1,.., lambda_n$ are the eigenvalues of $A$ and $beta_1,.., beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $sigma$ such that
$$tr(AB)=lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $lambda_1,.., lambda_n$ and $beta_1,.., beta_n$, (and a permutation $sigma$ which can actually be ignored by reordering $beta$) under which condition do we have
$$lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}leq lambda_1+..+lambda_n ?$$
Note that for any such $lambda, beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
$endgroup$
Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.
Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that
$$A=PD_1P^T \
B=PD_2P^{T}$$
Then
$$AB=PD_1D_2P^{T}$$
Therefore, if $lambda_1,.., lambda_n$ are the eigenvalues of $A$ and $beta_1,.., beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $sigma$ such that
$$tr(AB)=lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}$$
The question you are asking in this case is basically the following
Question Given some non-negative numbers $lambda_1,.., lambda_n$ and $beta_1,.., beta_n$, (and a permutation $sigma$ which can actually be ignored by reordering $beta$) under which condition do we have
$$lambda_1 beta_{sigma(1)}+...+lambda_n beta_{sigma(n)}leq lambda_1+..+lambda_n ?$$
Note that for any such $lambda, beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.
If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $leq 1$.
If $A,B$ don't commute, the question seems to be way to complicated.
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
answered Jan 20 at 18:03
N. S.N. S.
104k7114209
104k7114209
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
add a comment |
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
Thank you for the nice answer! Unfortunately, A and B don't commute, and I'm going to make more assumptions about their structure, but I think it's better to put it as a new question.
$endgroup$
– Babak
Jan 20 at 18:58
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
$begingroup$
I changed the problem assumptions as in math.stackexchange.com/questions/3081016/…
$endgroup$
– Babak
Jan 20 at 19:10
add a comment |
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$begingroup$
What exactly is a PSD matrix?
$endgroup$
– Matt Samuel
Jan 20 at 17:49
$begingroup$
Positieve semi-definite
$endgroup$
– James
Jan 20 at 17:52
$begingroup$
Do you know if $A,B$ commute? If they don't the problem is much harder, as $AB$ may not be symmetric.
$endgroup$
– N. S.
Jan 20 at 17:53