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A linear algebra proof that the inverse of a strictly ultrametric matrix is a strictly diagonally dominant Stieltjes matrix. (English) Zbl 0803.15020
An $$n \times n$$ Stieltjes matrix $$A=(a_{ij})$$ is defined as a real symmetric and positive definite matrix with $$a_{ij} \leq 0$$ for all $$i \neq j$$ $$(1 \leq i,j \leq n)$$. It is well known that every Stieltjes matrix has an inverse which is a nonsingular symmetric matrix with nonnegative entries. In this note the authors show, by giving a counterexample, that the converse of this statement fails in general to be true for $$n\geq 3$$, and give a simpler proof of a partially converse theorem proved by S. Martinez, G. Michon and J. San Martan [Inverses of strictly ultrametric matrices are of Stieltjes type. SIAM J. Matrix Anal. Appl., 15, No. 1, 98-106 (1994; Zbl 0798.15030)], that a strictly ultrametric matrix $$A=(a_{ij})$$ is nonsingular and its inverse $$A^{- 1} = (\alpha_{ij})$$ is a strictly diagonal dominant Stieltjes matrix with the additional property: $$\alpha_{ij} = 0$$ if and if $$a_{ij} = 0$$, using more familiar tools from linear algebras.

##### MSC:
 15B48 Positive matrices and their generalizations; cones of matrices 15B57 Hermitian, skew-Hermitian, and related matrices 15A09 Theory of matrix inversion and generalized inverses
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