Proof involving the spectral radius and Jordan Canonical form Announcing the arrival of Valued...

Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?

Why is black pepper both grey and black?

Why one of virtual NICs called bond0?

ListPlot join points by nearest neighbor rather than order

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

Gastric acid as a weapon

Output the ŋarâþ crîþ alphabet song without using (m)any letters

What does the "x" in "x86" represent?

Letter Boxed validator

What causes the vertical darker bands in my photo?

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

"Seemed to had" is it correct?

Is the Standard Deduction better than Itemized when both are the same amount?

Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?

Should I call the interviewer directly, if HR aren't responding?

3 doors, three guards, one stone

Why are there no cargo aircraft with "flying wing" design?

What makes black pepper strong or mild?

Why is "Captain Marvel" translated as male in Portugal?

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

Why aren't air breathing engines used as small first stages

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

The logistics of corpse disposal

How to draw this diagram using TikZ package?



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 $mathbb{Q}$












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. 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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 57 mins ago









      mXdXmXdX

      1018




      1018






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          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









          $endgroup$





















            1












            $begingroup$

            Hint



            $$A=PJP^{-1} \
            J=begin{bmatrix}
            lambda_1 & * & 0 & 0 & 0 & ... & 0 \
            0& lambda_2 & * & 0 & 0 & ... & 0 \
            ...&...&...&...&....&....&....\
            0 & 0 & 0 & 0&0&...&lambda_n \
            end{bmatrix}$$
            where each $*$ is either $0$ or $1$.



            Prove by induction that
            $$J^m=begin{bmatrix}
            lambda_1^m & star & star & star & star & ... & star \
            0& lambda_2^m & star & star & star & ... & star \
            ...&...&...&...&....&....&....\
            0 & 0 & 0 & 0&0&...&lambda_n^m \
            end{bmatrix}$$
            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, $A^{m} = PJ^{m}P^{-1}$. 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
              19 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.
              15 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
              9 mins 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-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









            4












            $begingroup$

            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









            $endgroup$


















              4












              $begingroup$

              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









              $endgroup$
















                4












                4








                4





                $begingroup$

                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 26 mins ago









                Robert IsraelRobert Israel

                332k23221478




                332k23221478























                    1












                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    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, $A^{m} = PJ^{m}P^{-1}$. 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
                      19 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.
                      15 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
                      9 mins ago
















                    1












                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    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, $A^{m} = PJ^{m}P^{-1}$. 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
                      19 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.
                      15 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
                      9 mins ago














                    1












                    1








                    1





                    $begingroup$

                    Hint



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    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



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    where each $*$ is either $0$ or $1$.



                    Prove by induction that
                    $$J^m=begin{bmatrix}
                    lambda_1^m & star & star & star & star & ... & star \
                    0& lambda_2^m & star & star & star & ... & star \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n^m \
                    end{bmatrix}$$
                    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 41 mins ago









                    N. S.N. S.

                    105k7115210




                    105k7115210












                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. 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
                      19 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.
                      15 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
                      9 mins ago


















                    • $begingroup$
                      So, $A^{m} = PJ^{m}P^{-1}$. 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
                      19 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.
                      15 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
                      9 mins ago
















                    $begingroup$
                    So, $A^{m} = PJ^{m}P^{-1}$. 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
                    19 mins ago




                    $begingroup$
                    So, $A^{m} = PJ^{m}P^{-1}$. 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
                    19 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.
                    15 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.
                    15 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
                    9 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
                    9 mins 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-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

                    ORA-01691 (unable to extend lob segment) even though my tablespace has AUTOEXTEND onORA-01692: unable to...

                    Image
                    Do I need to convey a moral for each of my blog post? Does convergence of polynomials imply that of its coefficients? Are hand made posters acceptable in Academia? How to read string as hex number in bash? Have the tides ever turned twice on any open problem? How do researchers send unsolicited emails asking for feedback on their works? How to determine the greatest d orbital splitting? When did hardware antialiasing start being available? How to test the sharpness of a knife? Asserting that Atheism and Theism are both faith based positions Print a physical multiplication table
                    Read more

                    Always On Availability groups resolving state after failover - Remote harden of transaction...

                    Image
                    Should corporate security training be tailored based on a users' job role? Does human life have innate value over that of other animals? Limit involving inverse functions Father gets chickenpox, but doesn't infect his two children. How is this possible? Is it possible to re-order a List of continous entries? Who, if anyone, was the first astronaut to return to earth in a different vessel? The totem pole can be grouped into Identical projects by students at two different colleges: still plagiarism? Sci fi book, man buys a beat up spaceship and intervenes in a civil war on a planet and eventually becomes a space cop How to not
                    Read more

                    Circunscripción electoral de Guipúzcoa Referencias Menú de navegaciónLas claves del sistema electoral en...

                    Image
                    Circunscripciones electorales del Senado de EspañaCircunscripciones electorales del Congreso de los DiputadosCircunscripciones electorales del Parlamento Vasco españolasCongreso de los DiputadosSenadobajaaltaPaís Vascoelecciones autonómicasprovincia homónima Localización de la circunscripción electoral de Guipúzcoa. Guipúzcoa es una de las 52 circunscripciones electorales españolas utilizadas como distritos electorales para el Congreso de los Diputados y el Senado, que son las cámaras baja y alta del Parlamento Español. [ 1 ] ​ Le corresponden 6 diputados y 4 senadores. También es una de las 3 circunscripciones electorales del País Vasco para las elecciones autonómicas, en que elige 25 parlamentarios. Se corresponde con la provincia homónima. [ 2 ] ​ Referencias ↑ Las claves del sistema electoral en España ↑ Resultados de las elecciones autonómicas de 2012 en Guipúzcoa This page is only for reference, If you need detailed information, please check here
                    Read more