Turning a square matrix of ones invertible by the least number of sign changes

Question

This is a follow up of my Previous question.
I am trying to prove the following claim (which I am not sure if is correct):
For any natural number $n\geq 2$, Transforming $J_n$ (square matrix whose entries are only $1$ 's) to an invertible matrix requires changing the sign of at least $n-1$ entries.
For example, the matrix $\begin{pmatrix}1 & 1\\ 1 &1\end{pmatrix}$ becomes invertible when changing one of the entries to $-1$, and obtaining the matrix $\begin{pmatrix}1 & 1\\ 1 &-1\end{pmatrix}$.

This was the base of the induction. I proceeded as follows:Assume the induction hypothesis holds for $n$, and prove for $n+1$. Assume by contradiction that $J_{n+1}$ can be turned into an invertible matrix by changing the sign of less than $n$ entries. Denote the number of negated entries by $k$. We have that $k<n<n+1$, which implies $k\leq n-1$, which implies that $n+1-k\geq 2$. Therefore, by the pigeon-hole principle, there must be at least 2 rows/columns of $J_{n+1}$ which were not changed. In other words, $J_{n+1}$ consists of at least 2 rows (or columns) of $1$'s, which implies that $J_{n+1}$ is singular. contradiction.
Feedback is appreciated.

Answer 1

Your proof is correct.

$n-1$ is a sharp lower bound. Simply change the elements along the main diagonal, except for the $(1,1)$ element which remains positive. Subtracting the first row from each of the other rows then gives a triangular matrix whose determinant will be $(-2)^{n-1}$.