Proof involving the spectral radius and the 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 operatorSpectral radius, second induced normContinuity of the spectral radiusSpectral radius and matrix norm inequality as its consequenceConfusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Spectral radius of a matrixLower bound spectral radius of matrixFinding the Jordan Form of a matrix…

How to deal with a team lead who never gives me credit?

What's the purpose of writing one's academic bio in 3rd person?

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

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

What are 'alternative tunings' of a guitar and why would you use them? Doesn't it make it more difficult to play?

How much radiation do nuclear physics experiments expose researchers to nowadays?

How can I fade player when goes inside or outside of the area?

Is there a concise way to say "all of the X, one of each"?

Why constant symbols in a language?

Antler Helmet: Can it work?

Why is black pepper both grey and black?

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

Why one of virtual NICs called bond0?

How widely used is the term Treppenwitz? Is it something that most Germans know?

Letter Boxed validator

Why is "Consequences inflicted." not a sentence?

Can Pao de Queijo, and similar foods, be kosher for Passover?

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

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

Are my PIs rude or am I just being too sensitive?

3 doors, three guards, one stone

Is there a "higher Segal conjecture"?

What LEGO pieces have "real-world" functionality?

List *all* the tuples!



Proof involving the spectral radius and the 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 operatorSpectral radius, second induced normContinuity of the spectral radiusSpectral radius and matrix norm inequality as its consequenceConfusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Spectral radius of a matrixLower bound spectral radius of matrixFinding the Jordan Form of a matrix…










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.










share|cite|improve this question











$endgroup$
















    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.










    share|cite|improve this question











    $endgroup$














      2












      2








      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.










      share|cite|improve this question











      $endgroup$





      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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 26 mins ago









      Rodrigo de Azevedo

      13.2k41961




      13.2k41961










      asked 1 hour ago









      mXdXmXdX

      1068




      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









          $endgroup$




















            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









            $endgroup$












            • $begingroup$
              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
              1 hour ago











            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


















            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















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            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









            $endgroup$

















              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









              $endgroup$















                5












                5








                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









                $endgroup$



                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 $A^n to 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Robert IsraelRobert Israel

                332k23221478




                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









                    $endgroup$












                    • $begingroup$
                      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
                      1 hour ago















                    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









                    $endgroup$












                    • $begingroup$
                      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
                      1 hour ago













                    2












                    2








                    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









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour 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
                      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
                      1 hour 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
                      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
                      1 hour 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
                    1 hour 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
                    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




                    $begingroup$
                    @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
                    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
                    1 hour ago

















                    draft saved

                    draft discarded
















































                    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.




                    draft saved


                    draft discarded














                    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















                    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







                    Popular posts from this blog

                    Disable / Remove link to Product Items in Cart Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How can I limit products that can be bought / added to cart?Remove item from cartHide “Add to Cart” button if specific products are already in cart“Prettifying” the custom options in cart pageCreate link in cart sidebar to view all added items After limit reachedLink products together in checkout/cartHow to Get product from cart and add it againHide action-edit on cart page if simple productRemoving Cart items - ObserverRemove wishlist items when added to cart

                    Helsingin valtaus Sisällysluettelo Taustaa | Yleistä sotatoimista | Osapuolet | Taistelut Helsingin ympäristössä | Punaisten antautumissuunnitelma | Taistelujen kulku Helsingissä | Valtauksen jälkeen | Tappiot | Muistaminen | Kirjallisuutta | Lähteet | Aiheesta muualla | NavigointivalikkoTeoksen verkkoversioTeoksen verkkoversioGoogle BooksSisällissota Helsingissä päättyi tasan 95 vuotta sittenSaksalaisten ylivoima jyräsi punaisen HelsinginSuomalaiset kuvaavat sotien jälkiä kaupungeissa – katso kuvat ja tarinat tutuilta kulmiltaHelsingin valtaus 90 vuotta sittenSaksalaiset valtasivat HelsinginHyökkäys HelsinkiinHelsingin valtaus 12.–13.4. 1918Saksalaiset käyttivät ihmiskilpiä Helsingin valtauksessa 1918Teoksen verkkoversioTeoksen verkkoversioSaksalaiset hyökkäävät Etelä-SuomeenTaistelut LeppävaarassaSotilaat ja taistelutLeppävaara 1918 huhtikuussa. KapinatarinaHelsingin taistelut 1918Saksalaisten voitonparaati HelsingissäHelsingin valtausta juhlittiinSaksalaisten Helsinki vuonna 1918Helsingin taistelussa kaatuneet valkokaartilaisetHelsinkiin haudatut taisteluissa kaatuneet punaiset12.4.1918 Helsingin valtauksessa saksalaiset apujoukot vapauttavat kaupunginVapaussodan muistomerkkejä Helsingissä ja pääkaupunkiseudullaCrescendo / Vuoden 1918 Kansalaissodan uhrien muistomerkkim

                    Adjektiivitarina Tarinan tekeminen | Esimerkki: ennen | Esimerkki: jälkeen | Navigointivalikko