Ttykuu

Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?

If a contract sometimes uses the wrong name, is it still valid?

How to draw this diagram using TikZ package?

What is the longest distance a 13th-level monk can jump while attacking on the same turn?

Proof involving the spectral radius and the Jordan canonical form

When is phishing education going too far?

Center align columns in table ignoring minus signs?

Why does Python start at index -1 when indexing a list from the end?

What happens to sewage if there is no river near by?

What is this single-engine low-wing propeller plane?

Letter Boxed validator

I am not a queen, who am I?

If Jon Snow became King of the Seven Kingdoms what would his regnal number be?

Is above average number of years spent on PhD considered a red flag in future academia or industry positions?

Is 1 ppb equal to 1 μg/kg?

Sorting numerically

Is there a "higher Segal conjecture"?

Can a non-EU citizen traveling with me come with me through the EU passport line?

What would be the ideal power source for a cybernetic eye?

Does accepting a pardon have any bearing on trying that person for the same crime in a sovereign jurisdiction?

Is there a documented rationale why the House Ways and Means chairman can demand tax info?

Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?

Does surprise arrest existing movement?

Why did the IBM 650 use bi-quinary?

Proof involving the spectral radius and the Jordan canonical form

2

$begingroup$

Let $A$ be a square matrix. Show that if $$lim_n to infty A^n = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.

Hint: Use the Jordan canonical form.

I am self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem — I don't know where to start. Any help appreciated.

linear-algebra matrices jordan-normal-form spectral-radius

share|cite|improve this question

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

$endgroup$

add a comment | 

2

$begingroup$

Let $A$ be a square matrix. Show that if $$lim_n to infty A^n = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.

Hint: Use the Jordan canonical form.

I am self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem — I don't know where to start. Any help appreciated.

linear-algebra matrices jordan-normal-form spectral-radius

share|cite|improve this question

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

$endgroup$

add a comment | 

$begingroup$

Let $A$ be a square matrix. Show that if $$lim_n to infty A^n = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.

Hint: Use the Jordan canonical form.

I am self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem — I don't know where to start. Any help appreciated.

linear-algebra matrices jordan-normal-form spectral-radius

share|cite|improve this question

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

$endgroup$

share|cite|improve this question

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

share|cite|improve this question

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

edited 23 mins ago

Rodrigo de Azevedo

13.2k41961

asked 1 hour ago

mXdXmXdX

1068

asked 1 hour ago

mXdXmXdX

1068

2 Answers
2

active

oldest

votes

5

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

2

$begingroup$

Hint

$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 1 hour ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^m = PJ^mP^-1$. If I can show what you're asking by induction, would the limit of $J^m = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^-1 A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^m$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  58 mins ago
  

  
  
  
  

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
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189376%2fproof-involving-the-spectral-radius-and-the-jordan-canonical-form%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

5

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

answered 1 hour ago

Robert IsraelRobert Israel

332k23221478

2

$begingroup$

Hint

$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 1 hour ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^m = PJ^mP^-1$. If I can show what you're asking by induction, would the limit of $J^m = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^-1 A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^m$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  58 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint

$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 1 hour ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^m = PJ^mP^-1$. If I can show what you're asking by induction, would the limit of $J^m = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^-1 A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^m$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  58 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint

$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 1 hour ago

N. S.N. S.

105k7115210

$endgroup$

share|cite|improve this answer

answered 1 hour ago

N. S.N. S.

105k7115210

answered 1 hour ago

N. S.N. S.

105k7115210

answered 1 hour ago

N. S.N. S.

105k7115210