Proof involving the spectral radius and Jordan Canonical form Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Spectral radius of the Volterra operatorExample that the Jordan canonical form is not “robust.”The unit vector in the direction of uWhat is the purpose of Jordan Canonical Form?Confusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixSpectral radius of perturbed bipartite graphsA proof involving invertible $ntimes n$ matricesProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Computing Canonical Jordan Form over a field $mathbbQ$
Why are there no cargo aircraft with "flying wing" design?
Is there a Spanish version of "dot your i's and cross your t's" that includes the letter 'ñ'?
What is the musical term for a note that continously plays through a melody?
Single word antonym of "flightless"
What causes the vertical darker bands in my photo?
Stars Make Stars
How can players work together to take actions that are otherwise impossible?
Why one of virtual NICs called bond0?
How do I keep my slimes from escaping their pens?
How discoverable are IPv6 addresses and AAAA names by potential attackers?
How do I stop a creek from eroding my steep embankment?
G-Code for resetting to 100% speed
Does surprise arrest existing movement?
The logistics of corpse disposal
When is phishing education going too far?
Is there a concise way to say "all of the X, one of each"?
Are my PIs rude or am I just being too sensitive?
What makes black pepper strong or mild?
Why does Python start at index -1 when indexing a list from the end?
Should I discuss the type of campaign with my players?
Doubts about chords
If a contract sometimes uses the wrong name, is it still valid?
Is it true to say that an hosting provider's DNS server is what links the entire hosting environment to ICANN?
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
Proof involving the spectral radius and Jordan Canonical form
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Spectral radius of the Volterra operatorExample that the Jordan canonical form is not “robust.”The unit vector in the direction of uWhat is the purpose of Jordan Canonical Form?Confusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixSpectral radius of perturbed bipartite graphsA proof involving invertible $ntimes n$ matricesProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Computing Canonical Jordan Form over a field $mathbbQ$
$begingroup$
Let $A$ be a square matrix. Show that if $lim_n to infty A^n = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.
I'm 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 spectral-radius
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 $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.
I'm 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 spectral-radius
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 $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.
I'm 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 spectral-radius
asked 1 hour ago
mXdXmXdX
1068
$endgroup$
Let $A$ be a square matrix. Show that if $lim_n to infty A^n = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.
I'm 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 spectral-radius
linear-algebra spectral-radius
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
1068
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $A^n to 0$.
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
$endgroup$
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.
answered 46 mins 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
25 mins 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.
20 mins 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
14 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
);
);
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%2f3189376%2fproof-involving-the-spectral-radius-and-jordan-canonical-form%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
$begingroup$
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 $A^n to 0$.
answered 31 mins 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 $A^n to 0$.
answered 31 mins 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 $A^n to 0$.
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
$endgroup$
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 $A^n to 0$.
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
answered 31 mins ago
Robert IsraelRobert Israel
332k23221478
332k23221478
add a comment |
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.
answered 46 mins 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
25 mins 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.
20 mins 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
14 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.
answered 46 mins 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
25 mins 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.
20 mins 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
14 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.
answered 46 mins ago
N. S.N. S.
105k7115210
$endgroup$
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.
answered 46 mins ago
N. S.N. S.
105k7115210
answered 46 mins ago
N. S.N. S.
105k7115210
answered 46 mins ago
N. S.N. S.
105k7115210
answered 46 mins ago
N. S.N. S.
105k7115210
105k7115210
$begingroup$
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
25 mins 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.
20 mins 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
14 mins ago
add a comment |
$begingroup$
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
25 mins 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.
20 mins 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
14 mins ago
$begingroup$
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
25 mins ago
$begingroup$
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
25 mins 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.
20 mins 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.
20 mins 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
14 mins 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
14 mins 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%2f3189376%2fproof-involving-the-spectral-radius-and-jordan-canonical-form%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
