If there a relationship between a submatrix's eigenvalues and the matrix's eigenvalues.











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My question




Let $A$ be a $n times n$ matrix. We denote the submatrix whose entries are ${a_{ij}}in A$ where $i=2,...,n$ and $j=2,...,n$ as $A_{n-1}$.



Suppose we know the eigenvalues of $A_{n-1}$. Does there exist a relationship between the eigenvalues of the matrix $A$ and the eigenvalues of the submatrix $A_{n-1}$?




Context



I am trying to do linear analysis on a system of $n$ ODEs an equilibrium point. The Jacobin of the system has the following structure
begin{align}
J&=begin{pmatrix}
-1 & a_{12} & dots & a_{1n} \
a_{21} & \
vdots & &A_{n-1}\
a_{n1} &
end{pmatrix}
end{align}
where $A_{n-1}$ is an $n-1 times n-1$ matrix whose eigenvalues I know from another unrelated method. The entries $a_{12},...,a_{1n}geq 0$ while entries $a_{21},...,a_{n1}leq 0$, note these entries are functions ans vary in value based on the equilibrium point the Jacobian is evaluated at.



Notes




  • I suspect that the answer to my question is that there is an upper bound for the eigenvalues of $A$. I haven't found anything that supports this suspicion.

  • I provide additional context to assist those answering the question, you can feel free to ignore the context if you can answer the original question.

  • If any clarification is needed feel free to ask.






matrices eigenvalues-eigenvectors jacobian







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    up vote
    0
    down vote

    favorite












    My question




    Let $A$ be a $n times n$ matrix. We denote the submatrix whose entries are ${a_{ij}}in A$ where $i=2,...,n$ and $j=2,...,n$ as $A_{n-1}$.



    Suppose we know the eigenvalues of $A_{n-1}$. Does there exist a relationship between the eigenvalues of the matrix $A$ and the eigenvalues of the submatrix $A_{n-1}$?




    Context



    I am trying to do linear analysis on a system of $n$ ODEs an equilibrium point. The Jacobin of the system has the following structure
    begin{align}
    J&=begin{pmatrix}
    -1 & a_{12} & dots & a_{1n} \
    a_{21} & \
    vdots & &A_{n-1}\
    a_{n1} &
    end{pmatrix}
    end{align}
    where $A_{n-1}$ is an $n-1 times n-1$ matrix whose eigenvalues I know from another unrelated method. The entries $a_{12},...,a_{1n}geq 0$ while entries $a_{21},...,a_{n1}leq 0$, note these entries are functions ans vary in value based on the equilibrium point the Jacobian is evaluated at.



    Notes




    • I suspect that the answer to my question is that there is an upper bound for the eigenvalues of $A$. I haven't found anything that supports this suspicion.

    • I provide additional context to assist those answering the question, you can feel free to ignore the context if you can answer the original question.

    • If any clarification is needed feel free to ask.






    matrices eigenvalues-eigenvectors jacobian







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      My question




      Let $A$ be a $n times n$ matrix. We denote the submatrix whose entries are ${a_{ij}}in A$ where $i=2,...,n$ and $j=2,...,n$ as $A_{n-1}$.



      Suppose we know the eigenvalues of $A_{n-1}$. Does there exist a relationship between the eigenvalues of the matrix $A$ and the eigenvalues of the submatrix $A_{n-1}$?




      Context



      I am trying to do linear analysis on a system of $n$ ODEs an equilibrium point. The Jacobin of the system has the following structure
      begin{align}
      J&=begin{pmatrix}
      -1 & a_{12} & dots & a_{1n} \
      a_{21} & \
      vdots & &A_{n-1}\
      a_{n1} &
      end{pmatrix}
      end{align}
      where $A_{n-1}$ is an $n-1 times n-1$ matrix whose eigenvalues I know from another unrelated method. The entries $a_{12},...,a_{1n}geq 0$ while entries $a_{21},...,a_{n1}leq 0$, note these entries are functions ans vary in value based on the equilibrium point the Jacobian is evaluated at.



      Notes




      • I suspect that the answer to my question is that there is an upper bound for the eigenvalues of $A$. I haven't found anything that supports this suspicion.

      • I provide additional context to assist those answering the question, you can feel free to ignore the context if you can answer the original question.

      • If any clarification is needed feel free to ask.






      matrices eigenvalues-eigenvectors jacobian







      share|cite|improve this question













      My question




      Let $A$ be a $n times n$ matrix. We denote the submatrix whose entries are ${a_{ij}}in A$ where $i=2,...,n$ and $j=2,...,n$ as $A_{n-1}$.



      Suppose we know the eigenvalues of $A_{n-1}$. Does there exist a relationship between the eigenvalues of the matrix $A$ and the eigenvalues of the submatrix $A_{n-1}$?




      Context



      I am trying to do linear analysis on a system of $n$ ODEs an equilibrium point. The Jacobin of the system has the following structure
      begin{align}
      J&=begin{pmatrix}
      -1 & a_{12} & dots & a_{1n} \
      a_{21} & \
      vdots & &A_{n-1}\
      a_{n1} &
      end{pmatrix}
      end{align}
      where $A_{n-1}$ is an $n-1 times n-1$ matrix whose eigenvalues I know from another unrelated method. The entries $a_{12},...,a_{1n}geq 0$ while entries $a_{21},...,a_{n1}leq 0$, note these entries are functions ans vary in value based on the equilibrium point the Jacobian is evaluated at.



      Notes




      • I suspect that the answer to my question is that there is an upper bound for the eigenvalues of $A$. I haven't found anything that supports this suspicion.

      • I provide additional context to assist those answering the question, you can feel free to ignore the context if you can answer the original question.

      • If any clarification is needed feel free to ask.






      matrices eigenvalues-eigenvectors jacobian




      matrices eigenvalues-eigenvectors jacobian






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      asked 2 days ago









      AzJ

      274118




      274118






















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          There is an answer to your question for symmetric matrices : the eigenvalues $lambda_k$ of $A_{n-1}$ and $mu_j$ of $A_n$ are interleaved :



          $$mu_1 leq lambda_1 leq mu_2 leq ... leq lambda_{n-1} leq mu_n$$



          This is a theorem established by Cauchy in the 1830s.



          See {Eigen values of a principal sub-matrix of a symmetric matrix}



          For non-symmetric matrices, there is no general result like this one.






          share|cite|improve this answer




























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            The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues ofbegin{bmatrix}-1&x\x&0end{bmatrix}are $dfrac{-1pmsqrt{1+4x^2}}2$, which can be arbitrarily large.






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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              There is an answer to your question for symmetric matrices : the eigenvalues $lambda_k$ of $A_{n-1}$ and $mu_j$ of $A_n$ are interleaved :



              $$mu_1 leq lambda_1 leq mu_2 leq ... leq lambda_{n-1} leq mu_n$$



              This is a theorem established by Cauchy in the 1830s.



              See {Eigen values of a principal sub-matrix of a symmetric matrix}



              For non-symmetric matrices, there is no general result like this one.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted










                There is an answer to your question for symmetric matrices : the eigenvalues $lambda_k$ of $A_{n-1}$ and $mu_j$ of $A_n$ are interleaved :



                $$mu_1 leq lambda_1 leq mu_2 leq ... leq lambda_{n-1} leq mu_n$$



                This is a theorem established by Cauchy in the 1830s.



                See {Eigen values of a principal sub-matrix of a symmetric matrix}



                For non-symmetric matrices, there is no general result like this one.






                share|cite|improve this answer























                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  There is an answer to your question for symmetric matrices : the eigenvalues $lambda_k$ of $A_{n-1}$ and $mu_j$ of $A_n$ are interleaved :



                  $$mu_1 leq lambda_1 leq mu_2 leq ... leq lambda_{n-1} leq mu_n$$



                  This is a theorem established by Cauchy in the 1830s.



                  See {Eigen values of a principal sub-matrix of a symmetric matrix}



                  For non-symmetric matrices, there is no general result like this one.






                  share|cite|improve this answer












                  There is an answer to your question for symmetric matrices : the eigenvalues $lambda_k$ of $A_{n-1}$ and $mu_j$ of $A_n$ are interleaved :



                  $$mu_1 leq lambda_1 leq mu_2 leq ... leq lambda_{n-1} leq mu_n$$



                  This is a theorem established by Cauchy in the 1830s.



                  See {Eigen values of a principal sub-matrix of a symmetric matrix}



                  For non-symmetric matrices, there is no general result like this one.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Jean Marie

                  28.1k41848




                  28.1k41848






















                      up vote
                      2
                      down vote













                      The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues ofbegin{bmatrix}-1&x\x&0end{bmatrix}are $dfrac{-1pmsqrt{1+4x^2}}2$, which can be arbitrarily large.






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues ofbegin{bmatrix}-1&x\x&0end{bmatrix}are $dfrac{-1pmsqrt{1+4x^2}}2$, which can be arbitrarily large.






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues ofbegin{bmatrix}-1&x\x&0end{bmatrix}are $dfrac{-1pmsqrt{1+4x^2}}2$, which can be arbitrarily large.






                          share|cite|improve this answer












                          The only eigenvalue if $[0]$ is, of course, $0$. However, the eigenvalues ofbegin{bmatrix}-1&x\x&0end{bmatrix}are $dfrac{-1pmsqrt{1+4x^2}}2$, which can be arbitrarily large.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 days ago









                          José Carlos Santos

                          140k18111204




                          140k18111204






























                               

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