Matrix rescaling increases lowest eigenvalue?

The matrices are of the form $$ A=\begin{pmatrix} 1 & C \\ C^* & D \end{pmatrix}, \quad\quad B= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} A \begin{pmatrix} 1 & 0\\ 0&2\end{pmatrix} , $$ with the blocks corresponding to the sizes of the sets $M_j$ involved.

Let $v=(x,y)^t$ be a normalized eigenvector for the minimum eigenvalue $\lambda $ of $B$. Then $$ (x,2y)A\begin{pmatrix} x \\ 2y \end{pmatrix} = \lambda $$ also, but this modified vector has larger norm.

So the desired inequality $\lambda_j(A)\le\lambda_j(B)$ (for all eigenvalues, not just the first one) will follow if we can show that $B\ge 0$. This is true for $y=1$ because in this case we can interpret $$ |M_j\cap M_k|=\sum_n \chi_j(n)\chi_k(n) $$ as the scalar product in $\ell^2$ of the characteristic functions, and this makes $v^*Bv$ equal to $\|f\|_2^2\ge 0$, with $f=\sum v_j\chi_j$.

For general $y>0$, we have $B(y)=(1/y)B(1) + D$, for a diagonal matrix $D$ with non-negative entries.