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Proof involving the spectral radius and the Jordan canonical form

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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…












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 34 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



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












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            2 Answers
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            2 Answers
            2






            active

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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 1 hour ago









                Robert IsraelRobert Israel

                332k23221478




                332k23221478























                    2












                    $begingroup$

                    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
                      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



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



                    $$A=PJP^{-1} \
                    J=begin{bmatrix}
                    lambda_1 & * & 0 & 0 & 0 & ... & 0 \
                    0& lambda_2 & * & 0 & 0 & ... & 0 \
                    ...&...&...&...&....&....&....\
                    0 & 0 & 0 & 0&0&...&lambda_n \
                    end{bmatrix}$$
                    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 1 hour 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
                      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, $A^{m} = PJ^{m}P^{-1}$. 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
                      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, $A^{m} = PJ^{m}P^{-1}$. 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
                    1 hour 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
                    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


















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