How to show the trace inequality of two P.S.D matrices $text{Tr(X)}leqtext{Tr(Y)}$ when $X preceq Y$?












1












$begingroup$


Let $X,Y$ be two Positive Semi-Definite matrices. How can we show the following in the most elegant and shortest way? Because I know how to prove it but I think there is a better way?
Alos, MaoWao shows it differently using summation.



$$text{Tr}(X)leqtext{Tr}(Y)$$ when $X preceq Y$, where $Y-X$ is positive semi-definite?



My try is:



$x^T(Y-X)x geq 0$ so $text{Tr}(xx^T(Y-X)) geq 0$.



Then using $text{tr}(AB) leq text{tr(A)} text{tr(B)}$



$$0 leq text{Tr}(xx^T(Y-X)) leq text{Tr}(xx^T)text{Tr}(Y-X)$$
hence the claim.



I do not want to use this $text{tr}(AB) leq text{tr(A)} text{tr(B)}$ or sumation. Is there any other way to show it shorter?






matrices inequality positive-definite







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:04










  • $begingroup$
    @ Lord Shark the Unknown: Yes
    $endgroup$
    – Saeed
    Jan 15 at 17:05






  • 1




    $begingroup$
    Then each diagonal entry in $Y-X$ is nonnegative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:07










  • $begingroup$
    @ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
    $endgroup$
    – Saeed
    Jan 15 at 17:39
















1












$begingroup$


Let $X,Y$ be two Positive Semi-Definite matrices. How can we show the following in the most elegant and shortest way? Because I know how to prove it but I think there is a better way?
Alos, MaoWao shows it differently using summation.



$$text{Tr}(X)leqtext{Tr}(Y)$$ when $X preceq Y$, where $Y-X$ is positive semi-definite?



My try is:



$x^T(Y-X)x geq 0$ so $text{Tr}(xx^T(Y-X)) geq 0$.



Then using $text{tr}(AB) leq text{tr(A)} text{tr(B)}$



$$0 leq text{Tr}(xx^T(Y-X)) leq text{Tr}(xx^T)text{Tr}(Y-X)$$
hence the claim.



I do not want to use this $text{tr}(AB) leq text{tr(A)} text{tr(B)}$ or sumation. Is there any other way to show it shorter?






matrices inequality positive-definite







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:04










  • $begingroup$
    @ Lord Shark the Unknown: Yes
    $endgroup$
    – Saeed
    Jan 15 at 17:05






  • 1




    $begingroup$
    Then each diagonal entry in $Y-X$ is nonnegative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:07










  • $begingroup$
    @ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
    $endgroup$
    – Saeed
    Jan 15 at 17:39














1












1








1





$begingroup$


Let $X,Y$ be two Positive Semi-Definite matrices. How can we show the following in the most elegant and shortest way? Because I know how to prove it but I think there is a better way?
Alos, MaoWao shows it differently using summation.



$$text{Tr}(X)leqtext{Tr}(Y)$$ when $X preceq Y$, where $Y-X$ is positive semi-definite?



My try is:



$x^T(Y-X)x geq 0$ so $text{Tr}(xx^T(Y-X)) geq 0$.



Then using $text{tr}(AB) leq text{tr(A)} text{tr(B)}$



$$0 leq text{Tr}(xx^T(Y-X)) leq text{Tr}(xx^T)text{Tr}(Y-X)$$
hence the claim.



I do not want to use this $text{tr}(AB) leq text{tr(A)} text{tr(B)}$ or sumation. Is there any other way to show it shorter?






matrices inequality positive-definite







share|cite|improve this question











$endgroup$




Let $X,Y$ be two Positive Semi-Definite matrices. How can we show the following in the most elegant and shortest way? Because I know how to prove it but I think there is a better way?
Alos, MaoWao shows it differently using summation.



$$text{Tr}(X)leqtext{Tr}(Y)$$ when $X preceq Y$, where $Y-X$ is positive semi-definite?



My try is:



$x^T(Y-X)x geq 0$ so $text{Tr}(xx^T(Y-X)) geq 0$.



Then using $text{tr}(AB) leq text{tr(A)} text{tr(B)}$



$$0 leq text{Tr}(xx^T(Y-X)) leq text{Tr}(xx^T)text{Tr}(Y-X)$$
hence the claim.



I do not want to use this $text{tr}(AB) leq text{tr(A)} text{tr(B)}$ or sumation. Is there any other way to show it shorter?






matrices inequality positive-definite




matrices inequality positive-definite






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 17:06







Saeed

















asked Jan 15 at 17:02









SaeedSaeed

1,171310




1,171310












  • $begingroup$
    What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:04










  • $begingroup$
    @ Lord Shark the Unknown: Yes
    $endgroup$
    – Saeed
    Jan 15 at 17:05






  • 1




    $begingroup$
    Then each diagonal entry in $Y-X$ is nonnegative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:07










  • $begingroup$
    @ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
    $endgroup$
    – Saeed
    Jan 15 at 17:39


















  • $begingroup$
    What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:04










  • $begingroup$
    @ Lord Shark the Unknown: Yes
    $endgroup$
    – Saeed
    Jan 15 at 17:05






  • 1




    $begingroup$
    Then each diagonal entry in $Y-X$ is nonnegative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 15 at 17:07










  • $begingroup$
    @ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
    $endgroup$
    – Saeed
    Jan 15 at 17:39
















$begingroup$
What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:04




$begingroup$
What does $X preceq Y$ mean? That $Y-X$ is positive semidefinite?
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:04












$begingroup$
@ Lord Shark the Unknown: Yes
$endgroup$
– Saeed
Jan 15 at 17:05




$begingroup$
@ Lord Shark the Unknown: Yes
$endgroup$
– Saeed
Jan 15 at 17:05




1




1




$begingroup$
Then each diagonal entry in $Y-X$ is nonnegative.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:07




$begingroup$
Then each diagonal entry in $Y-X$ is nonnegative.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 17:07












$begingroup$
@ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
$endgroup$
– Saeed
Jan 15 at 17:39




$begingroup$
@ Lord Shark the Unknown: You are right. Then to get the inequality we need use this fact and sum over the diagonals to get the result!
$endgroup$
– Saeed
Jan 15 at 17:39










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then
begin{align*}
e_i^intercal (Y - X) e_i = Y_{ii} - X_{ii} ge 0
end{align*}
Summing over $i$,
begin{align*}
text{tr}(Y-X) = sum_i(Y_{ii} - X_{ii}) ge 0
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
    $endgroup$
    – Saeed
    Jan 15 at 20:29










  • $begingroup$
    Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
    $endgroup$
    – Tom Chen
    Jan 16 at 5:29





















1












$begingroup$

Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) geq 0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
    $endgroup$
    – Saeed
    Jan 15 at 17:18










  • $begingroup$
    Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:28










  • $begingroup$
    @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
    $endgroup$
    – Saeed
    Jan 15 at 17:30










  • $begingroup$
    What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:50














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then
begin{align*}
e_i^intercal (Y - X) e_i = Y_{ii} - X_{ii} ge 0
end{align*}
Summing over $i$,
begin{align*}
text{tr}(Y-X) = sum_i(Y_{ii} - X_{ii}) ge 0
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
    $endgroup$
    – Saeed
    Jan 15 at 20:29










  • $begingroup$
    Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
    $endgroup$
    – Tom Chen
    Jan 16 at 5:29


















1












$begingroup$

Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then
begin{align*}
e_i^intercal (Y - X) e_i = Y_{ii} - X_{ii} ge 0
end{align*}
Summing over $i$,
begin{align*}
text{tr}(Y-X) = sum_i(Y_{ii} - X_{ii}) ge 0
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
    $endgroup$
    – Saeed
    Jan 15 at 20:29










  • $begingroup$
    Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
    $endgroup$
    – Tom Chen
    Jan 16 at 5:29
















1












1








1





$begingroup$

Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then
begin{align*}
e_i^intercal (Y - X) e_i = Y_{ii} - X_{ii} ge 0
end{align*}
Summing over $i$,
begin{align*}
text{tr}(Y-X) = sum_i(Y_{ii} - X_{ii}) ge 0
end{align*}






share|cite|improve this answer









$endgroup$



Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then
begin{align*}
e_i^intercal (Y - X) e_i = Y_{ii} - X_{ii} ge 0
end{align*}
Summing over $i$,
begin{align*}
text{tr}(Y-X) = sum_i(Y_{ii} - X_{ii}) ge 0
end{align*}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 18:23









Tom ChenTom Chen

2,143715




2,143715












  • $begingroup$
    still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
    $endgroup$
    – Saeed
    Jan 15 at 20:29










  • $begingroup$
    Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
    $endgroup$
    – Tom Chen
    Jan 16 at 5:29




















  • $begingroup$
    still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
    $endgroup$
    – Saeed
    Jan 15 at 20:29










  • $begingroup$
    Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
    $endgroup$
    – Tom Chen
    Jan 16 at 5:29


















$begingroup$
still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
$endgroup$
– Saeed
Jan 15 at 20:29




$begingroup$
still using summation! This is exactly what is in math.stackexchange.com/questions/3074592/… which I said. Can you show it without summation?
$endgroup$
– Saeed
Jan 15 at 20:29












$begingroup$
Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
$endgroup$
– Tom Chen
Jan 16 at 5:29






$begingroup$
Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares).
$endgroup$
– Tom Chen
Jan 16 at 5:29













1












$begingroup$

Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) geq 0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
    $endgroup$
    – Saeed
    Jan 15 at 17:18










  • $begingroup$
    Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:28










  • $begingroup$
    @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
    $endgroup$
    – Saeed
    Jan 15 at 17:30










  • $begingroup$
    What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:50


















1












$begingroup$

Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) geq 0.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
    $endgroup$
    – Saeed
    Jan 15 at 17:18










  • $begingroup$
    Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:28










  • $begingroup$
    @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
    $endgroup$
    – Saeed
    Jan 15 at 17:30










  • $begingroup$
    What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:50
















1












1








1





$begingroup$

Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) geq 0.$






share|cite|improve this answer









$endgroup$



Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) geq 0.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 17:10









OldGodzillaOldGodzilla

57427




57427












  • $begingroup$
    The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
    $endgroup$
    – Saeed
    Jan 15 at 17:18










  • $begingroup$
    Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:28










  • $begingroup$
    @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
    $endgroup$
    – Saeed
    Jan 15 at 17:30










  • $begingroup$
    What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:50




















  • $begingroup$
    The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
    $endgroup$
    – Saeed
    Jan 15 at 17:18










  • $begingroup$
    Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:28










  • $begingroup$
    @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
    $endgroup$
    – Saeed
    Jan 15 at 17:30










  • $begingroup$
    What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
    $endgroup$
    – OldGodzilla
    Jan 15 at 17:50


















$begingroup$
The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
$endgroup$
– Saeed
Jan 15 at 17:18




$begingroup$
The question is why all eigenvalues are nonnegative $text{Tr}(Y-X)leq 0$. Please read the statement carefully and confer to the provided links.
$endgroup$
– Saeed
Jan 15 at 17:18












$begingroup$
Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
$endgroup$
– OldGodzilla
Jan 15 at 17:28




$begingroup$
Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case).
$endgroup$
– OldGodzilla
Jan 15 at 17:28












$begingroup$
@Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
$endgroup$
– Saeed
Jan 15 at 17:30




$begingroup$
@Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer.
$endgroup$
– Saeed
Jan 15 at 17:30












$begingroup$
What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
$endgroup$
– OldGodzilla
Jan 15 at 17:50






$begingroup$
What is your definition of positive semidefinite? Is it that $x^* A x geq 0$ for all vectors $x neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^*$, where $U$ is unitary and $D$ is diagonal. Then $0 leq x^* A x = (Ux)^* D Ux = y^* D y$ for all vectors $y neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent.
$endgroup$
– OldGodzilla
Jan 15 at 17:50




















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