# Showing $u^T (M+I_n)^{-1} u \geq v^T(M+I_n)^{-1}v$ when $M$ is symmetric PD with positive entries and $u,v$ are $0-1$ vectors

Let $$n$$ be a positive integer. Let $$m_u,m_v \in \{1,...,n-1 \}$$.

Let $$M$$ be a $$n \times n$$ symmetric positive definite matrix with positive entries.

Let $$u$$ and $$v$$ be vectors of length $$n$$ with entries consisting $$n-m_u$$ (or $$n-m_v$$) $$0$$'s and $$m_u$$ (or $$m_v$$) $$1$$'s. Sort $$u$$ so that the first $$n-m_u$$ entires of $$u$$ are $$0$$'s and the last $$m_u$$ entries are $$1$$'s. Sort $$v$$ in the same way.

Suppose $$m_u>m_v$$. Is the following weak inequality true?

(As shown below by @Niki Di Giano, this is not true)

$$u^T (M+I_n)^{-1} u \geq v^T(M+I_n)^{-1}v$$

This is related to the post but now $$M$$ has positive entries only and we replace $$M$$ with $$(M+I_n)^{-1}$$.

Using almost the same matrix as the other post (if the matrix admits zero entries): $$M = \begin{bmatrix} 1 & 0 & 0\\ 0 & 10 & 9\\ 0 & 9 & 10\\ \end{bmatrix}$$ Then its inverse with the added identity is: $$(M + I)^{-1}= \begin{bmatrix} 1/2 & 0 & 0\\ 0 & 11/40 & -9/40\\ 0 & -9/40 & 11/40\\ \end{bmatrix}$$ This gives $$u^T M u = 4/40$$ and $$v^T M v = 11/40$$ for the vectors $$u = (0, 1, 1)$$ and $$v =(0, 0, 1)$$ respectively.
$$M = \begin{bmatrix} 1 & 1 & 1\\ 1 & 100 & 99\\ 1 & 99 & 100\\ \end{bmatrix} \\ \implies (M + I)^{-1} = \begin{bmatrix} 100/199 & -1/398 & -1/398\\ -1/398 & 201/796 & - 197/796\\ -1/398 & - 197/796 & 201/796\\ \end{bmatrix}$$ Which gives $$8/796$$ and $$201/796$$ for the same two vectors.
• Thank you so much. I am wondering if I can bound $\frac{1+u^T(M+I)^{-1} u}{1+v^T(M+I)^{-1} v}$ by $\frac{1+m_v}{1+m_u}$ and $\frac{1+m_u}{1+m_v}$. The question is posted here. Sorry to bother you again, but would you plz help me on this? – kx526 Aug 13 '19 at 15:39