An inequality of matrix normIs there a condition for the following consequence?Orthogonal Inner Product...
Hacking a Safe Lock after 3 tries
What did Alexander Pope mean by "Expletives their feeble Aid do join"?
What do Xenomorphs eat in the Alien series?
Can a druid choose the size of its wild shape beast?
If I can solve Sudoku can I solve Travelling Salesman Problem(TSP)? If yes, how?
How difficult is it to simply disable/disengage the MCAS on Boeing 737 Max 8 & 9 Aircraft?
Why does Bach not break the rules here?
Who is flying the vertibirds?
How to make healing in an exploration game interesting
My adviser wants to be the first author
How to simplify this time periods definition interface?
Is a party consisting of only a bard, a cleric, and a warlock functional long-term?
Instead of Universal Basic Income, why not Universal Basic NEEDS?
Did Ender ever learn that he killed Stilson and/or Bonzo?
Do I need to be arrogant to get ahead?
Are ETF trackers fundamentally better than individual stocks?
Does Mathematica reuse previous computations?
How to create the Curved texte?
Employee lack of ownership
Life insurance that covers only simultaneous/dual deaths
A limit with limit zero everywhere must be zero somewhere
A sequence that has integer values for prime indexes only:
Is it normal that my co-workers at a fitness company criticize my food choices?
AG Cluster db upgrade by vendor
An inequality of matrix norm
Is there a condition for the following consequence?Orthogonal Inner Product Proofprove change of basis matrix is unitarya matrix metricMatrix of non-degenerate product invertible?Prove that there is a $uin V$, such that $<u,v_i>$ is greater than zero, for every $i in {{1,..,m}}$.Inner product of dual basisColumn Spaces and SubsetsProve matrix inequality in inner product spaceInduced inner product on tensor powers.
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
$endgroup$
add a comment |
$begingroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
$endgroup$
Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.
Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$
I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!
linear-algebra matrices functional-analysis norm
linear-algebra matrices functional-analysis norm
edited 1 hour ago
Bernard
123k741116
123k741116
asked 2 hours ago
bbwbbw
51739
51739
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
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%2f3149889%2fan-inequality-of-matrix-norm%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
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
$endgroup$
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
add a comment |
$begingroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
$endgroup$
For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
$$
begin{array}{ll}
& ||U_1 U_2 - V_1 V_2||\ \
= & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
= &
|| ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
end{array}
$$
(The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
$$
|| ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
$$
A similar bound obtains for $||V_1 (U_2 - V_2) ||$.
This should give you enough "building blocks".:)
edited 1 hour ago
answered 1 hour ago
avsavs
3,424513
3,424513
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
add a comment |
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
$begingroup$
Thank you so much!
$endgroup$
– bbw
1 hour ago
1
1
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
$begingroup$
You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
$endgroup$
– avs
1 hour ago
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%2f3149889%2fan-inequality-of-matrix-norm%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