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Let $A$ be a non-negative square matrix. Normalize $A$ by its spectral radius $\sigma(A)$, and call it $A_2 = A/\sigma(A)$. Does this normalization preserve the ratio between the two largest eigenvalues of $A$? That is, is the ratio $\lambda_1/\lambda_2$ the same for $A$ and $A_2$, where $\lambda_1$ and $\lambda_2$ are the absolute values of the first and second largest eigenvalues of the two matrices?

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Of course. If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, then the eigenvalues of $\alpha A$ are $\alpha\lambda_1,\ldots,\alpha\lambda_n$.

This can be easily seen via de Schur decomposition.

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Schurly you don't need anything so fancy: if $A v = \lambda v$, then $\alpha A v = \alpha \lambda v$. – Robert Israel Nov 30 '12 at 19:17
Good point. I guess I don't think often enough about finite dimension... – Martin Argerami Nov 30 '12 at 19:18

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