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S Dec 11, 2015 at 11:43 comment added FFT The matrix $A = E -\epsilon I$, is a positive matrix for any $0<\epsilon <1$ and its eigenvalues are $n-\epsilon >0$ (simple Perron eigenvalue) and $-\epsilon$ with multiplicity $n-1$. So $A$ has $n-1$ negative eigenvalues.
S Dec 11, 2015 at 11:43 comment added FFT I agree with @Gottfried Helms, the set of totally positive matrices seems to be a subset of the set of entrywise nonnegative matrices. Thus I don't see why > positive matrices are just one side of a more general subject, that of Totally positive matrices. At the same time I don't see the relation between entrywise nonnegative (positive) matrices and nonnegative (or positive) eigenvalues, neither their real part (except, of course, for the spectral radius). For instance, consider the matrix $E$ made by all ones. (cont.)
Dec 10, 2015 at 21:55 comment added Gottfried Helms As I understand this the "totally positive matrix" are a subset of the OP's nonnegative matrices. So I think the qualification in this answer "...of a more general subject,..." is a bit misleading.
Dec 10, 2015 at 16:32 comment added Denis Serre @Surb. Precisely, that's the point. Total positivity means that $A$ is positive, and it induces positive matrices when acting on the exterior algebra. This implies that, not only $A$ has a positive real eigenvalue, but all its eigenvalues are positive (the converse is false).
Dec 10, 2015 at 14:25 comment added Surb Thank you for your answer. I'm not sure to understand how Positive matrices are only one side of [...] Totally positive matrices. It seems to me that $A=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ is a positive matrix which is not totally positive (its determinant is strictly negative).
Dec 10, 2015 at 13:20 history answered Denis Serre CC BY-SA 3.0