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Let $A \in \{\pm 1\}^{n \times n}$ be a symmetric matrix whose diagonal entries are $+1$. Let $f(A)$ be the smallest number of signs we need to change in $A$ so that it becomes positive semidefinite (while preserving symmetry). My questions are:

  1. Let $A_n$ be an $n \times n$ matrix with all off-diagonal entries equal to $-1$. What is the value of $f(A_n)$?

  2. Let $f_{\max}(n)$ be the maximum of $f(A)$ over all suitable $n \times n$ matrices. Is $f_{\max}(n) = f(A_n)$ true? What is the value of $f_{\max}(n)$?

I am also interested in the same questions for weighted $f(A)$, where changing any element by $x$ costs us $|x|$, and we want to make $A$ positive semidefinite as cheaply as possible.

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  • $\begingroup$ It will be worth checking if $f(A)$ is monotonic. That is, if $g_{i,j}(A)$ is the matrix resulting from switching the signs of $A_{i,j}$ and $A_{j,i}$ so that the smallest eigenvalue of $A$ is (strictly) increased, then does it follow that $f(g_{i,j}(A)) < f(A)$? $\endgroup$ – Josiah Park Nov 3 '18 at 15:47
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The answer to part 1 is $\lceil \frac{n^2}{2}\rceil-n$. Any fewer than that and the vector $v$ of all $1$'s will satisfy $v^\top Av< \lfloor\frac{n^2}{2}\rfloor-\lceil \frac{n^2}{2}\rceil\le 0$. Let $w$ be the $n$-by-$1$ vector with $i$ entry $(-1)^i$. Then $w w^\top$ is $\lceil \frac{n^2}{2}\rceil-n$ moves away from $A_n$. It has $n-1$ eigenvalues equal to $0$ and one eigenvalue equal to $n$ corresponding to the eigenvector $w$.

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