Polar decomposition of normal matrix
$begingroup$
Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.
I have no idea to prove this.
$A$ is normal then $AA^*=A^*A$. And then we have
$$
SS^*=U^*S^*SU.
$$
But I don't know how to continue.
linear-algebra
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
$endgroup$
add a comment |
$begingroup$
Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.
I have no idea to prove this.
$A$ is normal then $AA^*=A^*A$. And then we have
$$
SS^*=U^*S^*SU.
$$
But I don't know how to continue.
linear-algebra
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
$endgroup$
add a comment |
$begingroup$
Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.
I have no idea to prove this.
$A$ is normal then $AA^*=A^*A$. And then we have
$$
SS^*=U^*S^*SU.
$$
But I don't know how to continue.
linear-algebra
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
$endgroup$
Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.
I have no idea to prove this.
$A$ is normal then $AA^*=A^*A$. And then we have
$$
SS^*=U^*S^*SU.
$$
But I don't know how to continue.
linear-algebra
linear-algebra
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
edited Dec 13 '18 at 21:47
whereamI
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
asked Dec 13 '18 at 21:23
whereamIwhereamI
320115
320115
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $,A=U|A|$, then $,A^*=|A|U^*$. By normality one obtains
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $,U|A|U^* = |A|;Longleftrightarrow; U|A| = |A|U$.
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
$endgroup$
add a comment |
$begingroup$
Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU implies\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
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%2f3038582%2fpolar-decomposition-of-normal-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $,A=U|A|$, then $,A^*=|A|U^*$. By normality one obtains
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $,U|A|U^* = |A|;Longleftrightarrow; U|A| = |A|U$.
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
$endgroup$
add a comment |
$begingroup$
Let $,A=U|A|$, then $,A^*=|A|U^*$. By normality one obtains
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $,U|A|U^* = |A|;Longleftrightarrow; U|A| = |A|U$.
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
$endgroup$
add a comment |
$begingroup$
Let $,A=U|A|$, then $,A^*=|A|U^*$. By normality one obtains
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $,U|A|U^* = |A|;Longleftrightarrow; U|A| = |A|U$.
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
$endgroup$
Let $,A=U|A|$, then $,A^*=|A|U^*$. By normality one obtains
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $,U|A|U^* = |A|;Longleftrightarrow; U|A| = |A|U$.
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
answered Dec 13 '18 at 22:45
HannoHanno
2,223428
2,223428
add a comment |
add a comment |
$begingroup$
Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU implies\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU implies\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
add a comment |
$begingroup$
Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU implies\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU implies\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
answered Dec 13 '18 at 22:42
OmnomnomnomOmnomnomnom
128k791184
128k791184
add a comment |
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%2f3038582%2fpolar-decomposition-of-normal-matrix%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
