Show that $|Ax|_2^2 = lambda |x|_2^2$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Matrix norms involving singular valuesInequality matrix normDoes the spectral norm of a square matrix equal its largest eigenvalue in absolute value?Condition numbers and block matricescharacterization of the solution to a generalized eigenvalue problemProve $(lambda I-A)^-1=sum_i Big(fracv_iu_ilambda-lambda_iBig)$. $u_i, v_i$ are left and right eigenvectors for eigenvalue $lambda_i$Show that a matrix is invertible with norm less than one$A^tto 0$ when its row sum is strictly less than one?Absolute of all eigenvalues are always bounded by maximal singular valueShow that $ lambda leq |A^TA|$.
Why can't wing-mounted spoilers be used to steepen approaches?
Variable with quotation marks "$()"
Are there continuous functions who are the same in an interval but differ in at least one other point?
Do ℕ, mathbbN, BbbN, symbbN effectively differ, and is there a "canonical" specification of the naturals?
Example of compact Riemannian manifold with only one geodesic.
Why can't devices on different VLANs, but on the same subnet, communicate?
What aspect of planet Earth must be changed to prevent the industrial revolution?
What force causes entropy to increase?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
How did passengers keep warm on sail ships?
Simulating Exploding Dice
Could an empire control the whole planet with today's comunication methods?
How to handle characters who are more educated than the author?
different output for groups and groups USERNAME after adding a username to a group
Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?
Can I visit the Trinity College (Cambridge) library and see some of their rare books
Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)
Circular reasoning in L'Hopital's rule
Python - Fishing Simulator
Do I have Disadvantage attacking with an off-hand weapon?
How do you keep chess fun when your opponent constantly beats you?
60's-70's movie: home appliances revolting against the owners
Using dividends to reduce short term capital gains?
One-dimensional Japanese puzzle
Show that $|Ax|_2^2 = lambda |x|_2^2$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Matrix norms involving singular valuesInequality matrix normDoes the spectral norm of a square matrix equal its largest eigenvalue in absolute value?Condition numbers and block matricescharacterization of the solution to a generalized eigenvalue problemProve $(lambda I-A)^-1=sum_i Big(fracv_iu_ilambda-lambda_iBig)$. $u_i, v_i$ are left and right eigenvectors for eigenvalue $lambda_i$Show that a matrix is invertible with norm less than one$A^tto 0$ when its row sum is strictly less than one?Absolute of all eigenvalues are always bounded by maximal singular valueShow that $ lambda leq |A^TA|$.
$begingroup$
Let $A in mathbbR^n times n$, let $lambda$ be an eigenvalue of $A^TA$ and $x in mathbbR^n setminus 0$ be the corresponding eigenvector, then show that $$|Ax|_2^2 = lambda |x|_2^2 textand hence lambda geq 0$$
Answer:
Here $||.||_2$ denote matrix $ 2-$norm i.e, $||A||_2=sigma_max (A)=sqrtlambda,$ where $sigma_max$ is the largest singular value of matrix $A$ and $lambda$ is largest eigenvalue of $A^TA$.
Now we have,
$(A^TA)x=lambda x Rightarrow ||(A^TA)x||_2=||lambda x||_2$
How to conclude the proof?
help me.
Since
eigenvalues-eigenvectors norm matrix-norms
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
$endgroup$
|
show 5 more comments
$begingroup$
Let $A in mathbbR^n times n$, let $lambda$ be an eigenvalue of $A^TA$ and $x in mathbbR^n setminus 0$ be the corresponding eigenvector, then show that $$|Ax|_2^2 = lambda |x|_2^2 textand hence lambda geq 0$$
Answer:
Here $||.||_2$ denote matrix $ 2-$norm i.e, $||A||_2=sigma_max (A)=sqrtlambda,$ where $sigma_max$ is the largest singular value of matrix $A$ and $lambda$ is largest eigenvalue of $A^TA$.
Now we have,
$(A^TA)x=lambda x Rightarrow ||(A^TA)x||_2=||lambda x||_2$
How to conclude the proof?
help me.
Since
eigenvalues-eigenvectors norm matrix-norms
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
$endgroup$
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
1
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
1
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17
|
show 5 more comments
$begingroup$
Let $A in mathbbR^n times n$, let $lambda$ be an eigenvalue of $A^TA$ and $x in mathbbR^n setminus 0$ be the corresponding eigenvector, then show that $$|Ax|_2^2 = lambda |x|_2^2 textand hence lambda geq 0$$
Answer:
Here $||.||_2$ denote matrix $ 2-$norm i.e, $||A||_2=sigma_max (A)=sqrtlambda,$ where $sigma_max$ is the largest singular value of matrix $A$ and $lambda$ is largest eigenvalue of $A^TA$.
Now we have,
$(A^TA)x=lambda x Rightarrow ||(A^TA)x||_2=||lambda x||_2$
How to conclude the proof?
help me.
Since
eigenvalues-eigenvectors norm matrix-norms
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
$endgroup$
Let $A in mathbbR^n times n$, let $lambda$ be an eigenvalue of $A^TA$ and $x in mathbbR^n setminus 0$ be the corresponding eigenvector, then show that $$|Ax|_2^2 = lambda |x|_2^2 textand hence lambda geq 0$$
Answer:
Here $||.||_2$ denote matrix $ 2-$norm i.e, $||A||_2=sigma_max (A)=sqrtlambda,$ where $sigma_max$ is the largest singular value of matrix $A$ and $lambda$ is largest eigenvalue of $A^TA$.
Now we have,
$(A^TA)x=lambda x Rightarrow ||(A^TA)x||_2=||lambda x||_2$
How to conclude the proof?
help me.
Since
eigenvalues-eigenvectors norm matrix-norms
eigenvalues-eigenvectors norm matrix-norms
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
edited Mar 31 at 7:41
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
asked Mar 17 at 17:49
M. A. SARKARM. A. SARKAR
2,5001820
2,5001820
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
1
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
1
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17
|
show 5 more comments
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
1
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
1
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
1
1
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
1
1
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
A simpler proof is that $$||Ax||_2^2$$
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
Your Answer
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%2f3151828%2fshow-that-ax-22-lambda-x-22%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
$begingroup$
A simpler proof is that $$||Ax||_2^2$$
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
A simpler proof is that $$||Ax||_2^2$$
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
A simpler proof is that $$||Ax||_2^2$$
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
A simpler proof is that $$||Ax||_2^2$$
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 21:05
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
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%2f3151828%2fshow-that-ax-22-lambda-x-22%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
$begingroup$
Your 'answer' is not addressing the question asked. Use $|Ax|^2 = langle Ax, Ax rangle$.
$endgroup$
– copper.hat
Mar 17 at 17:52
1
$begingroup$
You've got that wrong. It's $left<Ax,Axright> = left<x, (A^TA)xright>$.
$endgroup$
– eyeballfrog
Mar 17 at 17:56
$begingroup$
@copper.hat, sorry, unfortunate. $||Ax||_2^2=<Ax,Ax>=leftlangle x,(A^TA)x rightrangle=leftlangle x, lambda x rightrangle=lambda <x,x>=lambda ||x||_2^2$
$endgroup$
– M. A. SARKAR
Mar 17 at 18:03
1
$begingroup$
@M.A.SARKAR: You got it!
$endgroup$
– copper.hat
Mar 17 at 18:05
$begingroup$
@copper.hat, How to show that $||A||_2 leq ||A^TA||^1/2$ ? , where $||.||$ is a norm on $mathbbR^n$.
$endgroup$
– M. A. SARKAR
Mar 17 at 18:17