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Let $A$ be a symmetric matrix and $A_i$ be the matrix obtained from $A$ by dropping the $i^{th}$ row and column. Then what are some good bounds on the value of $\frac{det(A_i)}{det(A)}$ ?

  • Using the fact that the eigenvalues of these two matrices interlace each other I can say that, $\frac{1}{\lambda_{max} (A) } \leq \frac{det(A_i)}{det(A)} \leq \frac{1}{\lambda_{min} (A) }$

    Can this be bettered? Feel free to assume something more about $A$ if you need!

I was wondering if one can somehow use the inequality that $\vert det (B) \vert \leq \prod_{i=1}^{i=dim(B)} \vert column_i (B) \vert _{l^2}$

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    $\begingroup$ You should assume A is nonsingular. Also, inspired by the idea of looking at a related ratio of the area of a face of a parallelipiped to its volume, I doubt that much more can be said. $\endgroup$
    – The Masked Avenger
    Commented Mar 13, 2015 at 20:18
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    $\begingroup$ Clearly, both bounds are sharp: take for $A$ a matrix whose minimal or maximal eigenvalue equals $1$ and the corresponding eigenspace splits off as an orthogonal summand. (I mean, is a coordinate line.) $\endgroup$
    – Alex Degtyarev
    Commented Mar 13, 2015 at 20:43
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    $\begingroup$ @AlexDegtyarev Simpler put: take a diagonal matrix. $\endgroup$
    – Igor Rivin
    Commented Mar 13, 2015 at 21:24
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    $\begingroup$ You can also use convexity of this function to get some bounds. $\endgroup$
    – Suvrit
    Commented Mar 13, 2015 at 22:09
  • $\begingroup$ Could you kindly elaborate? $\endgroup$
    – user6818
    Commented Mar 13, 2015 at 22:26

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