Mixing
What is the relation between $lVert{Hp}rVert^2$ and $lVert{H}rVert$ when $Hpne0$ and $lVert{p}rVert = 1$?
$begingroup$
What is the relation between $lVert{Hp}rVert^2$ and $lVert{H}rVert$ when $Hpne0$ and $lVert{p}rVert = 1$?
Note that $H$ and $p$ are sampled value of Gaussian distributed random variables, in which $H$ is a vector with a size of $1$ by $N$ and $p$ is a vector with a size of $N$ by $1$?
I think $$lVert{Hp}rVert^2 = lVert{H}rVert.$$
However, I failed to prove the above equation.
Is my thought wrong?
Thanks for reading my question.
vectors
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
$endgroup$
add a comment |
$begingroup$
What is the relation between $lVert{Hp}rVert^2$ and $lVert{H}rVert$ when $Hpne0$ and $lVert{p}rVert = 1$?
Note that $H$ and $p$ are sampled value of Gaussian distributed random variables, in which $H$ is a vector with a size of $1$ by $N$ and $p$ is a vector with a size of $N$ by $1$?
I think $$lVert{Hp}rVert^2 = lVert{H}rVert.$$
However, I failed to prove the above equation.
Is my thought wrong?
Thanks for reading my question.
vectors
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
$endgroup$
add a comment |
$begingroup$
What is the relation between $lVert{Hp}rVert^2$ and $lVert{H}rVert$ when $Hpne0$ and $lVert{p}rVert = 1$?
Note that $H$ and $p$ are sampled value of Gaussian distributed random variables, in which $H$ is a vector with a size of $1$ by $N$ and $p$ is a vector with a size of $N$ by $1$?
I think $$lVert{Hp}rVert^2 = lVert{H}rVert.$$
However, I failed to prove the above equation.
Is my thought wrong?
Thanks for reading my question.
vectors
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
$endgroup$
What is the relation between $lVert{Hp}rVert^2$ and $lVert{H}rVert$ when $Hpne0$ and $lVert{p}rVert = 1$?
Note that $H$ and $p$ are sampled value of Gaussian distributed random variables, in which $H$ is a vector with a size of $1$ by $N$ and $p$ is a vector with a size of $N$ by $1$?
I think $$lVert{Hp}rVert^2 = lVert{H}rVert.$$
However, I failed to prove the above equation.
Is my thought wrong?
Thanks for reading my question.
vectors
vectors
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
asked Jan 27 at 14:28
Danny_KimDanny_Kim
1,3911624
1,3911624
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
All you can state is
$$|Hp|^2 le |H|^2$$
Can you prove this one?
answered Jan 27 at 14:40
BerciBerci
61.7k23674
$endgroup$
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
add a 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%2f3089622%2fwhat-is-the-relation-between-lverthp-rvert2-and-lverth-rvert-when-hp%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
$begingroup$
All you can state is
$$|Hp|^2 le |H|^2$$
Can you prove this one?
answered Jan 27 at 14:40
BerciBerci
61.7k23674
$endgroup$
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
add a comment |
$begingroup$
All you can state is
$$|Hp|^2 le |H|^2$$
Can you prove this one?
answered Jan 27 at 14:40
BerciBerci
61.7k23674
$endgroup$
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
add a comment |
$begingroup$
All you can state is
$$|Hp|^2 le |H|^2$$
Can you prove this one?
answered Jan 27 at 14:40
BerciBerci
61.7k23674
$endgroup$
All you can state is
$$|Hp|^2 le |H|^2$$
Can you prove this one?
answered Jan 27 at 14:40
BerciBerci
61.7k23674
answered Jan 27 at 14:40
BerciBerci
61.7k23674
answered Jan 27 at 14:40
BerciBerci
61.7k23674
answered Jan 27 at 14:40
BerciBerci
61.7k23674
61.7k23674
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
add a comment |
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Actually, I tried many times using MATLAB program. I could observe that the value of RHS is much larger than that of LHS.
$endgroup$
– Danny_Kim
Jan 27 at 14:46
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
Well, by compactness of the unit sphere, there exists a unit vector $p$ such that $|Hp|=|H|$..
$endgroup$
– Berci
Jan 27 at 14:52
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
$begingroup$
However, to prove $lVert HprVert^2 le lVert HrVert^2$, do we have to prove that there do not exist $p$ such that $lVert HprVert^2 ge lVert HrVert^2$? Intuitively, each element of $p$ must be less than $1$, so each element of $Hp$ MAY be less than or equal to that of $H$. Is it enough to prove the above equation? I am not sure that it is enough.
$endgroup$
– Danny_Kim
Jan 27 at 14:55
add a 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%2f3089622%2fwhat-is-the-relation-between-lverthp-rvert2-and-lverth-rvert-when-hp%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
