When is the matrix norm multiplicative
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Let $|| = ||_p,q$ be an operator norm on $mathbb R^n times m$. In General, $|AB|le |A||B|$. Is there some criterion on $A, B$ (at least for some operator norms) so that $|AB| = lVert A rVert lVert BrVert$?
fa.functional-analysis linear-algebra matrix-theory
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
$endgroup$
add a comment |
$begingroup$
Let $|| = ||_p,q$ be an operator norm on $mathbb R^n times m$. In General, $|AB|le |A||B|$. Is there some criterion on $A, B$ (at least for some operator norms) so that $|AB| = lVert A rVert lVert BrVert$?
fa.functional-analysis linear-algebra matrix-theory
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
$endgroup$
6
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34
add a comment |
$begingroup$
Let $|| = ||_p,q$ be an operator norm on $mathbb R^n times m$. In General, $|AB|le |A||B|$. Is there some criterion on $A, B$ (at least for some operator norms) so that $|AB| = lVert A rVert lVert BrVert$?
fa.functional-analysis linear-algebra matrix-theory
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
$endgroup$
Let $|| = ||_p,q$ be an operator norm on $mathbb R^n times m$. In General, $|AB|le |A||B|$. Is there some criterion on $A, B$ (at least for some operator norms) so that $|AB| = lVert A rVert lVert BrVert$?
fa.functional-analysis linear-algebra matrix-theory
fa.functional-analysis linear-algebra matrix-theory
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
edited Feb 12 at 10:09
Martin Sleziak
3,08032231
3,08032231
asked Feb 12 at 8:32
N_SegolN_Segol
242
asked Feb 12 at 8:32
N_SegolN_Segol
242
asked Feb 12 at 8:32
N_SegolN_Segol
242
242
6
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34
add a comment |
6
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34
6
6
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$|x|=|x^*|=|xx^*|^1/2.$$
In the special case of $bf M_n(mathbb C)$, this is
$$|AA^*|=|A|cdot|A^*|$$
where the norm is the usual operator norm $|cdot|_2,2$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
$endgroup$
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
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: "504"
;
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%2fmathoverflow.net%2fquestions%2f323037%2fwhen-is-the-matrix-norm-multiplicative%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
$begingroup$
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$|x|=|x^*|=|xx^*|^1/2.$$
In the special case of $bf M_n(mathbb C)$, this is
$$|AA^*|=|A|cdot|A^*|$$
where the norm is the usual operator norm $|cdot|_2,2$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
$endgroup$
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
add a comment |
$begingroup$
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$|x|=|x^*|=|xx^*|^1/2.$$
In the special case of $bf M_n(mathbb C)$, this is
$$|AA^*|=|A|cdot|A^*|$$
where the norm is the usual operator norm $|cdot|_2,2$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
$endgroup$
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
add a comment |
$begingroup$
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$|x|=|x^*|=|xx^*|^1/2.$$
In the special case of $bf M_n(mathbb C)$, this is
$$|AA^*|=|A|cdot|A^*|$$
where the norm is the usual operator norm $|cdot|_2,2$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
$endgroup$
At least, there is this important case: in $C^*$-algebras, there is an involution and the norm has the property that
$$|x|=|x^*|=|xx^*|^1/2.$$
In the special case of $bf M_n(mathbb C)$, this is
$$|AA^*|=|A|cdot|A^*|$$
where the norm is the usual operator norm $|cdot|_2,2$.
Actually this formula extends to the case of rectangular matrices.
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
edited Feb 12 at 15:31
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
answered Feb 12 at 10:35
Denis SerreDenis Serre
29.3k795196
29.3k795196
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
add a comment |
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
$begingroup$
Thanks for the reply. In my case the matrices are usually not square. Do you know of a special family of matrices where $||AB|| = ||A||||B||$
$endgroup$
– N_Segol
Feb 12 at 12:00
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f323037%2fwhen-is-the-matrix-norm-multiplicative%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
6
$begingroup$
A remark: you cannot have a multiplicative norm on the whole $mathrmMat(n×m, , mathbbR)$, because there are non-zero matrices whose product is zero.
$endgroup$
– Francesco Polizzi
Feb 12 at 10:37
$begingroup$
@Francesco Polizzi unless $m = n = 1$...
$endgroup$
– KConrad
Feb 12 at 12:52
$begingroup$
@KConrad: of course, thank you.
$endgroup$
– Francesco Polizzi
Feb 12 at 13:44
$begingroup$
There is also the wide class of unitarily-invariant norms (i.e., norms depending only on singular values, like the operator norm, trace norm, Frobenius norm, Ky Fan norms, or Schatten norms). As their name suggests, the desired property holds for them if either of A or B is unitary.
$endgroup$
– Nathaniel Johnston
Feb 13 at 21:34