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

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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 $mathbb{Q}$












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$begingroup$


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









$endgroup$

















    2












    $begingroup$


    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









    $endgroup$















      2












      2








      2





      $begingroup$


      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 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
              31 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.
              26 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
              20 mins 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 37 mins 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
                      31 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.
                      26 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
                      20 mins 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
                      31 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.
                      26 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
                      20 mins 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 52 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
                      31 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.
                      26 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
                      20 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
                      31 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.
                      26 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
                      20 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
                    31 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
                    31 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.
                    26 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.
                    26 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
                    20 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
                    20 mins ago


















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