Is there any specific relationship among the determinant of leading principal submatrices of a tridiagonal matrix?

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The following symmetric matrix $A$ is given. Let denote $A_i$ the $i$-th leading principal submatrix of the matrix $A$.



$$
A = beginbmatrix
a_1 & b_1 \
b_1 & a_2 & b_2 \
& b_2 & ddots & ddots \
& & ddots & ddots & b_n-1 \
& & & b_n-1 & a_n
endbmatrix
$$



The determinant of the matrix $A$ is given by the recurrence relation



$det(A_i) = a_i det(A_i-1) - b_i-1^2det(A_i-2)$



I would like to know is there any other relation among the determinant of the leading principal submatrix of $A$? for example a numerically that determines determinant of $A_j$ is greater than $A_m$ for $j > m$ or any thing else.







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    up vote
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    The following symmetric matrix $A$ is given. Let denote $A_i$ the $i$-th leading principal submatrix of the matrix $A$.



    $$
    A = beginbmatrix
    a_1 & b_1 \
    b_1 & a_2 & b_2 \
    & b_2 & ddots & ddots \
    & & ddots & ddots & b_n-1 \
    & & & b_n-1 & a_n
    endbmatrix
    $$



    The determinant of the matrix $A$ is given by the recurrence relation



    $det(A_i) = a_i det(A_i-1) - b_i-1^2det(A_i-2)$



    I would like to know is there any other relation among the determinant of the leading principal submatrix of $A$? for example a numerically that determines determinant of $A_j$ is greater than $A_m$ for $j > m$ or any thing else.







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      The following symmetric matrix $A$ is given. Let denote $A_i$ the $i$-th leading principal submatrix of the matrix $A$.



      $$
      A = beginbmatrix
      a_1 & b_1 \
      b_1 & a_2 & b_2 \
      & b_2 & ddots & ddots \
      & & ddots & ddots & b_n-1 \
      & & & b_n-1 & a_n
      endbmatrix
      $$



      The determinant of the matrix $A$ is given by the recurrence relation



      $det(A_i) = a_i det(A_i-1) - b_i-1^2det(A_i-2)$



      I would like to know is there any other relation among the determinant of the leading principal submatrix of $A$? for example a numerically that determines determinant of $A_j$ is greater than $A_m$ for $j > m$ or any thing else.







      share|cite|improve this question














      The following symmetric matrix $A$ is given. Let denote $A_i$ the $i$-th leading principal submatrix of the matrix $A$.



      $$
      A = beginbmatrix
      a_1 & b_1 \
      b_1 & a_2 & b_2 \
      & b_2 & ddots & ddots \
      & & ddots & ddots & b_n-1 \
      & & & b_n-1 & a_n
      endbmatrix
      $$



      The determinant of the matrix $A$ is given by the recurrence relation



      $det(A_i) = a_i det(A_i-1) - b_i-1^2det(A_i-2)$



      I would like to know is there any other relation among the determinant of the leading principal submatrix of $A$? for example a numerically that determines determinant of $A_j$ is greater than $A_m$ for $j > m$ or any thing else.









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      edited Aug 8 at 18:32









      Bernard

      110k635103




      110k635103










      asked Aug 8 at 18:11









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