Mixing
How to show Von Neumann Trace inequality $ text{Tr}(AB) leq sum_{i=1}^n sigma_{A,i}sigma_{B,i} $?
$begingroup$
Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ text{Tr}(AB) leq sum_{i=1}^n sigma_{A,i}sigma_{B,i} $?
Also, what is the intuition behind this inequality?
matrices inequality trace
asked Jan 16 at 3:46
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ text{Tr}(AB) leq sum_{i=1}^n sigma_{A,i}sigma_{B,i} $?
Also, what is the intuition behind this inequality?
matrices inequality trace
asked Jan 16 at 3:46
SaeedSaeed
1,036310
$endgroup$
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
2
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26
add a comment |
$begingroup$
Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ text{Tr}(AB) leq sum_{i=1}^n sigma_{A,i}sigma_{B,i} $?
Also, what is the intuition behind this inequality?
matrices inequality trace
asked Jan 16 at 3:46
SaeedSaeed
1,036310
$endgroup$
Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ text{Tr}(AB) leq sum_{i=1}^n sigma_{A,i}sigma_{B,i} $?
Also, what is the intuition behind this inequality?
matrices inequality trace
matrices inequality trace
asked Jan 16 at 3:46
SaeedSaeed
1,036310
asked Jan 16 at 3:46
SaeedSaeed
1,036310
asked Jan 16 at 3:46
SaeedSaeed
1,036310
asked Jan 16 at 3:46
SaeedSaeed
1,036310
asked Jan 16 at 3:46
SaeedSaeed
1,036310
1,036310
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
2
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26
add a comment |
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
2
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
2
2
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is not a single, absolute "intuition" behind the inequality, because one can prove and interpret the inequality in different ways. In the following proof, one may say that the trace inequality is true because it is just a piling of a bunch of Cauchy-Schwarz inequalities.
Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$ to
$$
|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)tag{1}
$$
where $U,V$ are any two unitary matrices and $S=operatorname{diag}(s_1,ldots,s_n), D=operatorname{diag}(d_1,ldots,d_n)$ are any two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_koplus0_{n-k}$. Note that $S$ is a non-negatively weighted combinations of the $P_i$s. In fact,
$$
S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n
$$
and similarly for $D$. For convenience, let us write $S=sum_ka_kP_k$ and $D=sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes
$$
left|sum_{k,l}a_kb_loperatorname{tr}(UP_kV^ast P_l)right|
lesum_{k,l}a_kb_loperatorname{tr}(P_kP_l).tag{2}
$$
So, if we are able to prove that
$$
left|operatorname{tr}(UP_kV^ast P_l)right|
leoperatorname{tr}(P_kP_l)tag{3}
$$
for each pair of $k$ and $l$, then by triangle inequality, we obtain $(2)$ follows immediately.
To prove $(3)$, assume that $kle l$ (otherwise, interchange the roles of $k$ and $l$). Then the LHS of $(3)$ can be rewritten as $left|operatorname{tr}(UP_kV^ast P_l)right|=left|operatorname{tr}left((P_lVP_k)^ast (P_lUP_k)right)right|$ and hence $(3)$ becomes
$$
left|sum_{i=1}^k langle P_lu_i,,P_lv_irangleright|le k,tag{4}
$$
but this is obviously true because $P_l$ is an orthogonal projection and the columns of $U$ and $V$ are unit vectors.
answered Jan 16 at 19:38
user1551user1551
72.9k566128
$endgroup$
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
|
show 1 more comment
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075288%2fhow-to-show-von-neumann-trace-inequality-texttrab-leq-sum-i-1n-sigm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is not a single, absolute "intuition" behind the inequality, because one can prove and interpret the inequality in different ways. In the following proof, one may say that the trace inequality is true because it is just a piling of a bunch of Cauchy-Schwarz inequalities.
Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$ to
$$
|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)tag{1}
$$
where $U,V$ are any two unitary matrices and $S=operatorname{diag}(s_1,ldots,s_n), D=operatorname{diag}(d_1,ldots,d_n)$ are any two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_koplus0_{n-k}$. Note that $S$ is a non-negatively weighted combinations of the $P_i$s. In fact,
$$
S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n
$$
and similarly for $D$. For convenience, let us write $S=sum_ka_kP_k$ and $D=sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes
$$
left|sum_{k,l}a_kb_loperatorname{tr}(UP_kV^ast P_l)right|
lesum_{k,l}a_kb_loperatorname{tr}(P_kP_l).tag{2}
$$
So, if we are able to prove that
$$
left|operatorname{tr}(UP_kV^ast P_l)right|
leoperatorname{tr}(P_kP_l)tag{3}
$$
for each pair of $k$ and $l$, then by triangle inequality, we obtain $(2)$ follows immediately.
To prove $(3)$, assume that $kle l$ (otherwise, interchange the roles of $k$ and $l$). Then the LHS of $(3)$ can be rewritten as $left|operatorname{tr}(UP_kV^ast P_l)right|=left|operatorname{tr}left((P_lVP_k)^ast (P_lUP_k)right)right|$ and hence $(3)$ becomes
$$
left|sum_{i=1}^k langle P_lu_i,,P_lv_irangleright|le k,tag{4}
$$
but this is obviously true because $P_l$ is an orthogonal projection and the columns of $U$ and $V$ are unit vectors.
answered Jan 16 at 19:38
user1551user1551
72.9k566128
$endgroup$
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
|
show 1 more comment
$begingroup$
There is not a single, absolute "intuition" behind the inequality, because one can prove and interpret the inequality in different ways. In the following proof, one may say that the trace inequality is true because it is just a piling of a bunch of Cauchy-Schwarz inequalities.
Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$ to
$$
|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)tag{1}
$$
where $U,V$ are any two unitary matrices and $S=operatorname{diag}(s_1,ldots,s_n), D=operatorname{diag}(d_1,ldots,d_n)$ are any two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_koplus0_{n-k}$. Note that $S$ is a non-negatively weighted combinations of the $P_i$s. In fact,
$$
S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n
$$
and similarly for $D$. For convenience, let us write $S=sum_ka_kP_k$ and $D=sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes
$$
left|sum_{k,l}a_kb_loperatorname{tr}(UP_kV^ast P_l)right|
lesum_{k,l}a_kb_loperatorname{tr}(P_kP_l).tag{2}
$$
So, if we are able to prove that
$$
left|operatorname{tr}(UP_kV^ast P_l)right|
leoperatorname{tr}(P_kP_l)tag{3}
$$
for each pair of $k$ and $l$, then by triangle inequality, we obtain $(2)$ follows immediately.
To prove $(3)$, assume that $kle l$ (otherwise, interchange the roles of $k$ and $l$). Then the LHS of $(3)$ can be rewritten as $left|operatorname{tr}(UP_kV^ast P_l)right|=left|operatorname{tr}left((P_lVP_k)^ast (P_lUP_k)right)right|$ and hence $(3)$ becomes
$$
left|sum_{i=1}^k langle P_lu_i,,P_lv_irangleright|le k,tag{4}
$$
but this is obviously true because $P_l$ is an orthogonal projection and the columns of $U$ and $V$ are unit vectors.
answered Jan 16 at 19:38
user1551user1551
72.9k566128
$endgroup$
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
|
show 1 more comment
$begingroup$
There is not a single, absolute "intuition" behind the inequality, because one can prove and interpret the inequality in different ways. In the following proof, one may say that the trace inequality is true because it is just a piling of a bunch of Cauchy-Schwarz inequalities.
Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$ to
$$
|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)tag{1}
$$
where $U,V$ are any two unitary matrices and $S=operatorname{diag}(s_1,ldots,s_n), D=operatorname{diag}(d_1,ldots,d_n)$ are any two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_koplus0_{n-k}$. Note that $S$ is a non-negatively weighted combinations of the $P_i$s. In fact,
$$
S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n
$$
and similarly for $D$. For convenience, let us write $S=sum_ka_kP_k$ and $D=sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes
$$
left|sum_{k,l}a_kb_loperatorname{tr}(UP_kV^ast P_l)right|
lesum_{k,l}a_kb_loperatorname{tr}(P_kP_l).tag{2}
$$
So, if we are able to prove that
$$
left|operatorname{tr}(UP_kV^ast P_l)right|
leoperatorname{tr}(P_kP_l)tag{3}
$$
for each pair of $k$ and $l$, then by triangle inequality, we obtain $(2)$ follows immediately.
To prove $(3)$, assume that $kle l$ (otherwise, interchange the roles of $k$ and $l$). Then the LHS of $(3)$ can be rewritten as $left|operatorname{tr}(UP_kV^ast P_l)right|=left|operatorname{tr}left((P_lVP_k)^ast (P_lUP_k)right)right|$ and hence $(3)$ becomes
$$
left|sum_{i=1}^k langle P_lu_i,,P_lv_irangleright|le k,tag{4}
$$
but this is obviously true because $P_l$ is an orthogonal projection and the columns of $U$ and $V$ are unit vectors.
answered Jan 16 at 19:38
user1551user1551
72.9k566128
$endgroup$
There is not a single, absolute "intuition" behind the inequality, because one can prove and interpret the inequality in different ways. In the following proof, one may say that the trace inequality is true because it is just a piling of a bunch of Cauchy-Schwarz inequalities.
Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$ to
$$
|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)tag{1}
$$
where $U,V$ are any two unitary matrices and $S=operatorname{diag}(s_1,ldots,s_n), D=operatorname{diag}(d_1,ldots,d_n)$ are any two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_koplus0_{n-k}$. Note that $S$ is a non-negatively weighted combinations of the $P_i$s. In fact,
$$
S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n
$$
and similarly for $D$. For convenience, let us write $S=sum_ka_kP_k$ and $D=sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes
$$
left|sum_{k,l}a_kb_loperatorname{tr}(UP_kV^ast P_l)right|
lesum_{k,l}a_kb_loperatorname{tr}(P_kP_l).tag{2}
$$
So, if we are able to prove that
$$
left|operatorname{tr}(UP_kV^ast P_l)right|
leoperatorname{tr}(P_kP_l)tag{3}
$$
for each pair of $k$ and $l$, then by triangle inequality, we obtain $(2)$ follows immediately.
To prove $(3)$, assume that $kle l$ (otherwise, interchange the roles of $k$ and $l$). Then the LHS of $(3)$ can be rewritten as $left|operatorname{tr}(UP_kV^ast P_l)right|=left|operatorname{tr}left((P_lVP_k)^ast (P_lUP_k)right)right|$ and hence $(3)$ becomes
$$
left|sum_{i=1}^k langle P_lu_i,,P_lv_irangleright|le k,tag{4}
$$
but this is obviously true because $P_l$ is an orthogonal projection and the columns of $U$ and $V$ are unit vectors.
answered Jan 16 at 19:38
user1551user1551
72.9k566128
edited Jan 16 at 22:50
answered Jan 16 at 19:38
user1551user1551
72.9k566128
answered Jan 16 at 19:38
user1551user1551
72.9k566128
answered Jan 16 at 19:38
user1551user1551
72.9k566128
72.9k566128
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
|
show 1 more comment
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
Why $operatorname{tr}(USV^TD)leoperatorname{tr}(SD)$ is true? Could you prove this?
$endgroup$
– Saeed
Jan 16 at 22:03
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I do not understand how you have changed the Von Nuemann to your trace inequality. I cannot see the steps? Could you elaborate that?
$endgroup$
– Saeed
Jan 16 at 22:11
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
I stuck at this point: "Using the tracial property and singular value decomposition, one can reduce the inequality", How? I mean, How can you get this $|operatorname{tr}(USV^ast D)|leoperatorname{tr}(SD)$ from this $|operatorname{tr}(AB)|lesum_isigma_i(A)sigma_i(B)$? Let $A=U_ASV_A^ast$ and $B=U_BDV_B^ast$, so $|operatorname{tr}(U_ASV_A^ast U_BDV_B^ast)|=|operatorname{tr}(V_B^ast U_ASV_A^ast U_BD)|$. Is $U=V_B^ast U_A$ and $V=U_B^ast V_A$. Is that what you mean?
$endgroup$
– Saeed
Jan 16 at 22:45
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
Where does this come from? $S=(s_1-s_2)P_1+cdots+(s_{n-1}-s_n)P_{n-1}+s_nP_n$
$endgroup$
– Saeed
Jan 16 at 22:52
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
$begingroup$
@ user1551: Ok. Please explain it to me later.
$endgroup$
– Saeed
Jan 16 at 23:00
|
show 1 more comment
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075288%2fhow-to-show-von-neumann-trace-inequality-texttrab-leq-sum-i-1n-sigm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You need absolute values around the trace.
$endgroup$
– copper.hat
Jan 16 at 5:45
2
$begingroup$
See this: link.springer.com/article/10.1007/BF01647331
$endgroup$
– i707107
Jan 16 at 6:26