0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.