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 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 $mathbbQ$

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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 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 $mathbbQ$










2












$begingroup$



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



Hint: Use the Jordan canonical form.




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






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







share|cite|improve this question











$endgroup$
















    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.






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







    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.






      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 4 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
              49 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.
              44 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
              38 mins ago











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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

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            active

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            active

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            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 55 mins 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
                      49 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.
                      44 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
                      38 mins 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
                      49 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.
                      44 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
                      38 mins 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
                      49 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.
                      44 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
                      38 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
                      49 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.
                      44 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
                      38 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
                    49 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
                    49 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.
                    44 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.
                    44 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
                    38 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
                    38 mins ago

















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