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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 $\mathbb{R}$)?

Rodrigo de Azevedo
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2 Answers2

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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 $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.

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 $0$.

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 $\begin{pmatrix} 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}$

John Habert
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  • OK, I think your eigenvalues are $\tfrac{1}{2}\left(n\pm\sqrt{n^2-4}\right)$, which is bounded below by $1/n$. Note that this doesn't work for 2x2; my derivation starts with at least a 3x3. – Michael Grant Feb 07 '14 at 03:38
  • @MichaelC.Grant Added best so far for $n=4$ case. I think I've found a smaller but need to double check things. Even with the smaller one, bound of $\frac{1}{n+1}$ seems to be holding so far. – John Habert Feb 07 '14 at 03:42
  • I verified the correctness of my formula in the previous comment up to $n=1000$, for a matrix with all ones except a single zero in the $(1,2)$ position. That's not to say better matrices do not exist; but for a single off-diagonal nonzero, that's it. – Michael Grant Feb 07 '14 at 04:22
  • @MichaelC.Grant Nice work. I'm going to finish rechecking $n=4$ to see if the better case still exists. Discovered my code wasn't giving correct output. – John Habert Feb 07 '14 at 04:32
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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 $n=1000$.

Michael Grant
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  • Why doesn't your formula and my empirical result agree? – John Habert Feb 06 '14 at 22:03
  • Our matrices are different. Yours has one zero value, mine has $n-1$. – Michael Grant Feb 06 '14 at 22:06
  • Plus it appears I've miscalculated something. Double-checking on my own side shows my 3 by 3 answer is off. Will figure out why. – John Habert Feb 06 '14 at 22:08
  • Your matrix is $\vec{1}\vec{1}-e_1e_1^T$. Simpler than mine, to be sure. – Michael Grant Feb 06 '14 at 22:09
  • If it can be shown that the smallest absolute value is $1/n$ that would be great. –  Feb 06 '14 at 22:12
  • It's not going to be exactly $1/n$, obviously; it can be verified exhaustively for small matrices. – Michael Grant Feb 06 '14 at 22:15
  • Fixed my calculation for $n=3$. Verified exhaustively. I wonder if min is $|\frac{1}{2}((2n-3)-\sqrt{5})|$. I'll try later to exhaust $n=4$. – John Habert Feb 06 '14 at 22:21
  • I don't see the point in worrying about individual cases. The interesting question is the limiting behavior, and whether these are the best bounds possible. Your matrix produces a larger answer anyway. – Michael Grant Feb 06 '14 at 22:28
  • I'm hoping the individual cases will lead to a pattern which makes find the bounds much easier. Minimum eigenvalue from your described matrix is $|1-\sqrt{2}| \approx 0.414$. Minimum from mine is $\approx 0.3819$. – John Habert Feb 07 '14 at 01:11
  • Ah, nevermind, I haven't been doing an off-diagonal zero. You are absolutely right. Let me see if I can update my analysis with your matrix pattern. – Michael Grant Feb 07 '14 at 02:36