This is a follow-up question to this post: link . In general, given $n$, two matrices are defined as follows: \begin{equation} A = \begin{pmatrix} I_{n-m_A} & 0 \\ 0 & I_{m_A} + J_{m_A} \\ \end{pmatrix}, B = \begin{pmatrix} I_{n-m_B} & 0 \\ 0 & I_{m_B} + J_{m_B} \\ \end{pmatrix}, \end{equation} where $m_A \ne m_B$ and they can be $1,...,n-1$ (so it can be that $m_A < m_B$). $J_m$ is a $m \times m$ matrix of ones.

By the post, I know that the multiplicity of 1 as an eigenvalue is always $n-2$. Now, is it possible to show that $\prod_{i=1}^n\lambda_i(B^{-1}A) = O(1)$ as $n \to \infty$?

In Matlab, this seems to be true for all values of $m$. By the previously mentioned post, this product equals to the product of the two remaining eigenvalues: $\lambda_{1^*}(B^{-1}A) \times \lambda_{2^*}(B^{-1}A)$. I don't have a proof, but in Matlab, this seems to equal $\lambda_{1}(B^{-1}A) \times \lambda_{n}(B^{-1}A)$ (i.e. the product of the smallest and the largest eigenvalues).

  • $\begingroup$ What do you mean by $"=O(1)$ ? That the product of eigenvalues (i.e. the determinant) can be found in constant time (therefore independent from $n$) ? $\endgroup$ – Jean Marie Aug 8 at 16:11
  • $\begingroup$ @JeanMarie I want to show that there is a constant $M>0$, independent of $n$, such that $\prod_{i=1}^n \lambda_i < M$ when $n$ is large (or even better if I can show this for all $n$). $\endgroup$ – kx526 Aug 8 at 16:20

The product of the eigenvalues is the determinant, so what you're looking to calculate is $\det(B^{-1}A)=\frac{\det A}{\det B}$. Since $\det A = \det I_{n-m_A}\det(I_{m_A}+J_{m_A}) = \det(I_{m_A}+J_{m_A})$ and similarly for $\det B$, this ratio can depend only on $m_A$ and $m_B$ and not on $n$.

It's clear that $m+1$ is an eigenvalue of $I_m+J_m$ (with the vector of all $1$s as the corresponding eigenvector) and that $1$ is an eigenvalue of multiplicity $m-1$ (since $I_m+J_m-1I_m=J_m$ has rank $1$ and therefore nullity $m-1$). This then exhausts all the eigenvalues, so the product of the eigenvalues (and thus $\det(I_m+J_m)$) is $m+1$.

Then $\det(B^{-1}A)=\frac{\det A}{\det B} = \frac{m_A+1}{m_B+1}$. This can become arbitrarily small as $n\to\infty$ if $m_A$ stays small but $m_B$ grows, or conversely arbitrarily large if $m_A$ grows and $m_B$ stays small.

  • $\begingroup$ Thank you again :) $\endgroup$ – kx526 Aug 8 at 17:06
  • $\begingroup$ Sorry to bother you once more; I have a related question that I tried to solve using your approach above but has not been successful. It would be appreciated if you could share some inputs. Thank you. $\endgroup$ – kx526 Aug 9 at 15:39

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