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Are there relationships that quantify how the spectral radius of a matrix with only (0,1) entries behaves as some of the 1's get flipped to -1? Obviously with all entries flipped you have the same (absolute value) max eigenvalue, but I'm struggling to derive anything explicit for this. This seems like it shouldn't be too hard but I'm still struggling.

Also, how would you generalize that to complex entries with $|A_{ij}|\leq 1$?

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2 Answers 2

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The spectral radius of the modified matrix will never become larger than the original one. Gelfand's formula states that $\rho(A)=\lim_{k\to\infty}\|A^k\|^{1/k}$, so the assertion follows if you take the Frobenius norm.

The same reasoning still works if a nonnegative matrix is multiplied entrywise by a complex matrix whose elements lie on the closed unit disc.

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  • $\begingroup$ Could you please elaborate a bit? It's not obvious to me why the Frobenius norm demonstrates this. $\endgroup$
    – TeeJay
    Sep 6, 2017 at 6:34
  • $\begingroup$ @TeeJay Let $A$ be an entrywise nonnegative matrix. Multiply some of its entries (individually) by some complex numbers of magnitudes $\le1$ to obtain a new matrix $B$. Then you can prove by mathematical induction that the all entries of $B^k$ are bounded above by those of $A^k$ in magnitudes. Hence $\|A^k\|_F\ge\|B^k\|_F$. Now apply Gelfand's formula to conclude. $\endgroup$
    – user1551
    Sep 6, 2017 at 6:45
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There are no real obvious answers here; I think that the best we could do is get some kind of inequality.

Here's something we can say: let $A$ denote the $(0,1)$ matrix, let $B$ denote the version with some entries "flipped". Let $E = B - A$ (which has $-2$'s and $0$ for entries), and let $\|\cdot\|$ denote the spectral norm (distinct from the spectral radius; we could actually use any "matrix-norm" here).

We then have the quick inequality $$ \rho(B) \leq \|B\| = \|A + E\| \leq \|A\| + \|E\| $$

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