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  1. Free, publicly-accessible full text available June 1, 2023
  2. The classic graphical Cheeger inequalities state that if $M$ is an $n\times n$ \emph{symmetric} doubly stochastic matrix, then \[ \frac{1-\lambda_{2}(M)}{2}\leq\phi(M)\leq\sqrt{2\cdot(1-\lambda_{2}(M))} \] where $\phi(M)=\min_{S\subseteq[n],|S|\leq n/2}\left(\frac{1}{|S|}\sum_{i\in S,j\not\in S}M_{i,j}\right)$ is the edge expansion of $M$, and $\lambda_{2}(M)$ is the second largest eigenvalue of $M$. We study the relationship between $\phi(A)$ and the spectral gap $1-\re\lambda_{2}(A)$ for \emph{any} doubly stochastic matrix $A$ (not necessarily symmetric), where $\lambda_{2}(A)$ is a nontrivial eigenvalue of $A$ with maximum real part. Fiedler showed that the upper bound on $\phi(A)$ is unaffected, i.e., $\phi(A)\leq\sqrt{2\cdot(1-\re\lambda_{2}(A))}$. With regards to the lower bound on $\phi(A)$, there are known constructions with \[ \phi(A)\in\Theta\left(\frac{1-\re\lambda_{2}(A)}{\log n}\right), \] indicating that at least a mild dependence on $n$ is necessary to lower bound $\phi(A)$. In our first result, we provide an \emph{exponentially} better construction of $n\times n$ doubly stochastic matrices $A_{n}$, for which \[ \phi(A_{n})\leq\frac{1-\re\lambda_{2}(A_{n})}{\sqrt{n}}. \] In fact, \emph{all} nontrivial eigenvalues of our matrices are $0$, even though the matrices are highly \emph{nonexpanding}. We further show that this bound is in the correct range (up to the exponent of $n$), by showing that for any doubly stochastic matrix $A$, \[ \phi(A)\geq\frac{1-\re\lambda_{2}(A)}{35\cdot n}. \] As a consequence, unlike the symmetric case, there is a (necessary) loss of amore »factor of $n^{\alpha}$ for $\frac{1}{2}\leq\alpha\leq1$ in lower bounding $\phi$ by the spectral gap in the nonsymmetric setting. Our second result extends these bounds to general matrices $R$ with nonnegative entries, to obtain a two-sided \emph{gapped} refinement of the Perron-Frobenius theorem. Recall from the Perron-Frobenius theorem that for such $R$, there is a nonnegative eigenvalue $r$ such that all eigenvalues of $R$ lie within the closed disk of radius $r$ about $0$. Further, if $R$ is irreducible, which means $\phi(R)>0$ (for suitably defined $\phi$), then $r$ is positive and all other eigenvalues lie within the \textit{open} disk, so (with eigenvalues sorted by real part), $\re\lambda_{2}(R)« less