Consider the $n\times n$ matrix $F$ defined as $$ F=A-\varepsilon B $$ where $A$ is a constant matrix such that $a_{ij}=a>0$ for all $i,j$ and where $B$ is a symmetric matrix such that $b_{ij}\geq 0$ for all $i,j$.
I would like to find sufficient conditions on $B$ such that $F$ is positive definite when the scalar $\varepsilon>0$ is small enough.
Over at Mathoverflow Fedor Petrov shows that $F$ is positive semidefinite for $\varepsilon$ small enough if $B$ is positive definite on the subspace $H:\sum_i x_i=0$. It seems that a similar argument should works to show that $F$ is positive definite. I am probably missing something but can't figure out what.
Edit: Following Fedor's arguments, here is how I think the proof would go:
Since $B$ is negative definite on $H$, we have that $(Bx,x)\leqslant -c\|x\|^2$ for $x\in H$. We can write any vector $z$ as $z=x+y$ where $x\in H$ and $y \perp H$. Then \begin{align}((A-\varepsilon B)z,z)&=n\|y\|^2-\varepsilon (Bx,x)-2\varepsilon (Bx,y)-\varepsilon (By,y)\\ &\geqslant (n-1/2)\|y\|^2+c\varepsilon \|x\|^2-2\cdot \varepsilon \cdot \|B\|\cdot\|x\|\cdot\|y\|\end{align} for $\varepsilon>0$ small enough. This last expression is strictly convex in $\left(\|x\|,\|y\|\right)$ when $\varepsilon>0$ is small enough and reaches it minimum of 0 at $\left(\|x\|,\|y\|\right)=(0,0)$ or, equivalently, at $z=0$. Since $((A-\varepsilon B)z,z)>0$ for any $z\neq 0$, $((A-\varepsilon B)z,z)$ is positive definite.
Is there anything wrong with this proof?
The sufficient condition given in that MO answer is already one that makes $F$ positive definite for small $\varepsilon>0$.
In general, suppose $A$ and $B$ are two Hermitian matrices of the same sizes (whether they are entrywise positive/nonnegative is unimportant). If $A-\varepsilon B$ is positive semidefinite for every small $\varepsilon>0$, then by passing $\varepsilon$ to the limit, $A$ must be positive semidefinite. If $A$ is indeed positive semidefinite, denote its restriction on $(\ker A)^\perp$ as $A_1$. Then $A_1$ is positive definite and we may write $F=A-\varepsilon B$ in the form of $$ F=\pmatrix{A_1-\varepsilon X&-\varepsilon Y^\ast\\ -\varepsilon Y&-\varepsilon Z}. $$ Therefore, a necessary condition for $F$ to be positive semidefinite for every small $\varepsilon>0$ is that $A$ is positive semidefinite and $Z=(I-A^+A)B(I-A^+A)$ is negative semidefinite on $\ker A$.
However, if $A$ is positive semidefinite but $Z$ is negative definite on $\ker A$, we get a sufficient condition. This should be evident, because $F$ is congruent to $(-\varepsilon Z)\oplus S$, where \begin{align} S&=A_1-\varepsilon X-(-\varepsilon Y^\ast)(-\varepsilon Z)^{-1}(-\varepsilon Y)\\ &=A_1-\varepsilon (X-Y^\ast Z^{-1}Y) \end{align} is the Schur complement of $-\varepsilon Z$ in $F$. Since $A_1$ is positive definite and $Z$ is negative definite, $-\varepsilon Z,\ S,\ (-\varepsilon Z)\oplus S$ and in turn $F$ are positive definite when $\varepsilon>0$ is small.
In your case, $A$ is a positive multiple of the all-one matrix. So, $A$ is positive semidefinite, $\ker A=\{x\in\mathbb R^n:\sum_ix_i=0\}$ and $I-A^+A$, the orthogonal projection on $\ker A$, is the matrix $P$ in the MO answer you mentioned.