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Suppose I have an arbitrary real matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ such that the sum of absolute values in each row is greater than $1$:

$$\sum_{j=1}^n |A_{ij}|>1,\quad \forall i=1,\ldots,n.$$

(I guess this means that the infinity norm $||\mathbf{A}||_{\infty}$ of $\mathbf{A}$ is greater than $1$.)

Intuitively, this should imply that the spectral radius (largest eigenvalue in absolute terms) of $\mathbf{A}$ is greater than $1$. Is this true? And if so, how to prove it?

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Try $$ A=\pmatrix{1&-1\\1&-1}. $$

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