Smallest non-zero eigenvalue of a (0,1) matrix

Question

What's the smallest absolute value possible of a non-zero eigenvalue of an $n$ by $n$ square matrix whose entries are either $0$ or $1$ (all operations are over $\mathbb{R}$)?

Answer 1

For a $2\times 2$ matrix, the smallest is $\left| \dfrac{1-\sqrt{5}}{2}\right|$ from $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.

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For a $3\times 3$ matrix, the smallest is $\left|\frac{1}{2}(3-\sqrt{5})\right|$ from $\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ or any other matrix with all entries $1$ except for a single off (main) diagonal $0$.

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Best so far - have more calculations to run and check

For a $4\times 4$ matrix, the smallest is $\left|2-\sqrt{3}\right|$ from $\begin{pmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

Answer 2

Consider the $n\times n$ matrix $$E_n = \vec{1} \vec{1}^T + e_1 e_1^T - I$$ where $\vec{1}$ is the vector of all ones, $e_1$ is the vector with a $1$ in the first element and zeros elsewhere, and $I$ is the identity matrix. In other words, $E_n$ has ones everywhere except the latter $n-1$ elements of the diagonal.

Empirically, I'm finding that the smallest nonzero eigenvalue in absolute value is approximately $-1/n$. I suspect that could be bounded rigorously, and if I can do so, I'll edit this answer. But it would seem clear to me that the smallest non-zero eigenvalue cannot be bounded away from zero.

EDIT: The eigenvalues of $E_n$ for $n>2$ are $-1$ and $$\frac{n-1\pm\sqrt{(n-1)^2+4}}{2}=\frac{n-1}{2}\pm\sqrt{\left(\frac{n-1}{2}\right)^2+1}.$$ The smallest absolute value is therefore $$\sqrt{\left(\frac{n-1}{2}\right)^2+1}-\frac{n-1}{2}\geq \frac{1}{n-1}.$$ Of course, this is not a bound for all $\{0,1\}$ matrices, just for this one.

EDIT: John Habert's 3x3 matrices and 4x4 matrices do better than this. For a matrix will all ones except a single off-diagonal zero, the eigenvalues are 0 (with $n-2$ multiplicty) and $$\frac{n}{2} \pm \sqrt{ \frac{n^2}{4} - 1 } \geq \frac{1}{n}.$$ I verified this numerically up to $n=1000$.