Mixing
How to show $text{Tr}(AB) leq text{Tr}(AC)$ where $B preceq C$?
$begingroup$
Given three positive semi-definite matrices $A, B, C$. Show $operatorname{Tr}(AB) leq operatorname{Tr}(AC)$ where $B preceq C$?
This inequality is the matrix form of multiplying a positive number to both sides of an equality.
My attempt:
Since $A$ is P.S.D $A=A^{1/2}A^{1/2}$ so $text{Tr}(AB)=text{Tr}(A^{1/2}BA^{1/2})$, I need to show $text{Tr}(A^{1/2}BA^{1/2}) leq text{Tr}(A^{1/2}CA^{1/2})$ using $B preceq C$ which I stuck.
Also, if you can show it differently please add that method as well but please first complete my answer.
matrices inequality positive-semidefinite
edited Jan 15 at 17:00
darij grinberg
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Given three positive semi-definite matrices $A, B, C$. Show $operatorname{Tr}(AB) leq operatorname{Tr}(AC)$ where $B preceq C$?
This inequality is the matrix form of multiplying a positive number to both sides of an equality.
My attempt:
Since $A$ is P.S.D $A=A^{1/2}A^{1/2}$ so $text{Tr}(AB)=text{Tr}(A^{1/2}BA^{1/2})$, I need to show $text{Tr}(A^{1/2}BA^{1/2}) leq text{Tr}(A^{1/2}CA^{1/2})$ using $B preceq C$ which I stuck.
Also, if you can show it differently please add that method as well but please first complete my answer.
matrices inequality positive-semidefinite
edited Jan 15 at 17:00
darij grinberg
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Given three positive semi-definite matrices $A, B, C$. Show $operatorname{Tr}(AB) leq operatorname{Tr}(AC)$ where $B preceq C$?
This inequality is the matrix form of multiplying a positive number to both sides of an equality.
My attempt:
Since $A$ is P.S.D $A=A^{1/2}A^{1/2}$ so $text{Tr}(AB)=text{Tr}(A^{1/2}BA^{1/2})$, I need to show $text{Tr}(A^{1/2}BA^{1/2}) leq text{Tr}(A^{1/2}CA^{1/2})$ using $B preceq C$ which I stuck.
Also, if you can show it differently please add that method as well but please first complete my answer.
matrices inequality positive-semidefinite
edited Jan 15 at 17:00
darij grinberg
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
$endgroup$
Given three positive semi-definite matrices $A, B, C$. Show $operatorname{Tr}(AB) leq operatorname{Tr}(AC)$ where $B preceq C$?
This inequality is the matrix form of multiplying a positive number to both sides of an equality.
My attempt:
Since $A$ is P.S.D $A=A^{1/2}A^{1/2}$ so $text{Tr}(AB)=text{Tr}(A^{1/2}BA^{1/2})$, I need to show $text{Tr}(A^{1/2}BA^{1/2}) leq text{Tr}(A^{1/2}CA^{1/2})$ using $B preceq C$ which I stuck.
Also, if you can show it differently please add that method as well but please first complete my answer.
matrices inequality positive-semidefinite
matrices inequality positive-semidefinite
edited Jan 15 at 17:00
darij grinberg
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
edited Jan 15 at 17:00
darij grinberg
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
edited Jan 15 at 17:00
darij grinberg
11.3k33164
edited Jan 15 at 17:00
darij grinberg
11.3k33164
edited Jan 15 at 17:00
darij grinberg
11.3k33164
11.3k33164
asked Jan 15 at 16:12
SaeedSaeed
1,036310
asked Jan 15 at 16:12
SaeedSaeed
1,036310
asked Jan 15 at 16:12
SaeedSaeed
1,036310
1,036310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your first step is good. Since
$$
x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x,
$$
we have $A^{1/2}BA^{1/2}preceq A^{1/2}CA^{1/2}$. Thus
$$
mathrm{Tr}(A^{1/2}BA^{1/2})=sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_kleq sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=mathrm{Tr}(A^{1/2}CA^{1/2}),
$$
where $(e_1,dots,e_n)$ is an orthonormal basis of $mathbb{C}^n$.
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
$endgroup$
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your first step is good. Since
$$
x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x,
$$
we have $A^{1/2}BA^{1/2}preceq A^{1/2}CA^{1/2}$. Thus
$$
mathrm{Tr}(A^{1/2}BA^{1/2})=sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_kleq sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=mathrm{Tr}(A^{1/2}CA^{1/2}),
$$
where $(e_1,dots,e_n)$ is an orthonormal basis of $mathbb{C}^n$.
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
$endgroup$
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
add a comment |
$begingroup$
Your first step is good. Since
$$
x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x,
$$
we have $A^{1/2}BA^{1/2}preceq A^{1/2}CA^{1/2}$. Thus
$$
mathrm{Tr}(A^{1/2}BA^{1/2})=sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_kleq sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=mathrm{Tr}(A^{1/2}CA^{1/2}),
$$
where $(e_1,dots,e_n)$ is an orthonormal basis of $mathbb{C}^n$.
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
$endgroup$
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
add a comment |
$begingroup$
Your first step is good. Since
$$
x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x,
$$
we have $A^{1/2}BA^{1/2}preceq A^{1/2}CA^{1/2}$. Thus
$$
mathrm{Tr}(A^{1/2}BA^{1/2})=sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_kleq sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=mathrm{Tr}(A^{1/2}CA^{1/2}),
$$
where $(e_1,dots,e_n)$ is an orthonormal basis of $mathbb{C}^n$.
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
$endgroup$
Your first step is good. Since
$$
x^H A^{1/2}BA^{1/2} x=(A^{1/2} x)^HB(A^{1/2} x)leq (A^{1/2}x)^H C(A^{1/2} x)=x^{H}A^{1/2}C A^{1/2} x,
$$
we have $A^{1/2}BA^{1/2}preceq A^{1/2}CA^{1/2}$. Thus
$$
mathrm{Tr}(A^{1/2}BA^{1/2})=sum_{k=1}^n e_k^H A^{1/2}BA^{1/2} e_kleq sum_{k=1}^n e_k^H A^{1/2}CA^{1/2} e_k=mathrm{Tr}(A^{1/2}CA^{1/2}),
$$
where $(e_1,dots,e_n)$ is an orthonormal basis of $mathbb{C}^n$.
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
answered Jan 15 at 16:22
MaoWaoMaoWao
3,153617
3,153617
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
add a comment |
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Could you show it differently that $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?
$endgroup$
– Saeed
Jan 15 at 16:39
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
Well, if you know that a psd matrix has nonnegative eigenvalues, then you can use $0preceq Y-X$ to conclude $mathrm{Tr}(Y-X)geq 0$.
$endgroup$
– MaoWao
Jan 15 at 16:43
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
This is exactly my question. I know this. How can I get the last inequality in a shortest way without summation?
$endgroup$
– Saeed
Jan 15 at 16:46
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
$begingroup$
What is your definition of the trace? If it's the sum of eigenvalues, then see my last comment. If it's the sum of diagonal entries, then see my answer. In any case, there is gonna be some sum involved.
$endgroup$
– MaoWao
Jan 15 at 16:48
add a comment |
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