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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…












4












$begingroup$



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











$endgroup$












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago
















4












$begingroup$



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











$endgroup$












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago














4












4








4





$begingroup$



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











$endgroup$





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




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 2 hours ago









Rodrigo de Azevedo

13.2k41961




13.2k41961










asked 4 hours ago









mXdXmXdX

1168




1168












  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago


















  • $begingroup$
    This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
    $endgroup$
    – copper.hat
    1 hour ago
















$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
1 hour ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

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












    $begingroup$

    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
      3 hours 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.
      3 hours 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
      3 hours ago












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    8












    $begingroup$

    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









    $endgroup$


















      8












      $begingroup$

      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









      $endgroup$
















        8












        8








        8





        $begingroup$

        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 3 hours ago









        Robert IsraelRobert Israel

        332k23221478




        332k23221478























            4












            $begingroup$

            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
              3 hours 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.
              3 hours 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
              3 hours ago
















            4












            $begingroup$

            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
              3 hours 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.
              3 hours 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
              3 hours ago














            4












            4








            4





            $begingroup$

            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 3 hours 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
              3 hours 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.
              3 hours 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
              3 hours 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
              3 hours 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.
              3 hours 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
              3 hours 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
            3 hours 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
            3 hours 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.
            3 hours 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.
            3 hours 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
            3 hours 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
            3 hours ago


















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            Articole de calitateNașteri în 1871Nașteri pe 31 martieDecese în 1931Decese pe 15 februarieArtiști români din secolul al XIX-leaArtiști români din secolul al XX-leaCaricaturiști româniTinerimea artistică CraiovaRomâniaBucureștiRomâniacaricaturistȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiNicolae Vermont1898Honoré DaumierParlamentului României19081913Constantin Jiquidi1898secolului al XIX-leaGolemo KonjariBulgariaMuzeului de Artă din Craiova2007Craiova31 martie1871limba francezălimba latinăLiceul Sfântul SavaBucureștigazului metanOperei RomâneȘtefan Petrescu-MuscăLiceul „Carol I”Școlii de Belle-ArteCostin PetrescuGheorghe TattarescuTheodor AmanConstantin I. StăncescuAlexandru ObedenaruSalonului Artiștilor IndependențiExpoziția Artiștilor în ViațăNicolae VermontȘtefan LuchianConstantin ArtachinoAlexandru Bogdan-PiteștiIon AngelescuNicolae GrantIon TicuLudovic DolinskiCaricatura Eu Sunt... Școala de Belle-Arte din BucureștiAteneului RomânGeorge RanettiCafeneaua ImperialFi...
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