# Showing $\prod_{i=1}^n \lambda_i = O(1)$ for certain $n \times n$ matrix when 1 as an eigenvalue has multiplicity $n-2$

This is a follow-up question to this post: link . In general, given $$n$$, two matrices are defined as follows: $$$$A = \begin{pmatrix} I_{n-m_A} & 0 \\ 0 & I_{m_A} + J_{m_A} \\ \end{pmatrix}, B = \begin{pmatrix} I_{n-m_B} & 0 \\ 0 & I_{m_B} + J_{m_B} \\ \end{pmatrix},$$$$ where $$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).

• 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 $n$) ? – Jean Marie Aug 8 at 16:11
• @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 $n$). – 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.