Tricky question in Matrices!

Question

Define $$A=[\text{I}+\sum_{k=1}^{m}u_{k}u_{k}^T]^{-1}$$, where for each $u_k$ is a $0-1$ column vector.

Prove that for every $1\leq k \leq m$ $$Au_{k}u_{k}^T\geq0$$ i.e. each entry of $Au_ku_k^T$ is non-negative.

Answer 1

A randomly generated example:

Let $$ U=[u_1,u_2,u_3]= \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ Then $$ A = (I+UU^T)^{-1}= \begin{bmatrix} 0.5 & 0 & 0 & 0 & 0 \\ 0 & 0.625 & -0.25 & 0.125 & 0 \\ 0 & -0.25 & 0.5 & -0.25 & 0 \\ 0 & 0.125 & -0.25 & 0.625 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$ and $$ Au_3u_3^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -0.125 & -0.125 & 0 \\ 0 & 0 & 0.25 & 0.25 & 0 \\ 0 & 0 & 0.375 & 0.375 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \not\geq 0. $$

Answer 2

Take $u_1=\begin{bmatrix}1\\0\end{bmatrix}$, $u_2=\begin{bmatrix}1\\1\end{bmatrix}$, then $A=\frac{1}{5}\begin{bmatrix}2 & -1\\-1& 3\end{bmatrix}$ and $Au_1 u_1^T=\frac{1}{5}\begin{bmatrix}2 & 0\\-1 &0\end{bmatrix}$.