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If a matrix $A \in \mathbb{R}^{N\times N}$ is symmetric, tridiagonal, diagonally dominant, and all the diagonal elements of $A$ are positive, then is $A$ also positive-definite? If it is, how to prove it?

The proposition looks to be true according to a statement in this question, Is a symmetric positive definite matrix always diagonally dominant? and a comment to it. However, I can't come up to a proof by myself. I would appreciate your help.

norio
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    The matrix $-I$ is symmetric, diagonal (not just tridiagonal), and diagonally dominant. But it is not positive definite. Do you have any conditions on the individual entries? – JimmyK4542 Jul 13 '18 at 07:21
  • Right. It looks I need to add a condition that the diagonal elements of A are all positive, according to the statement in the cited question. I am going to edit my question. – norio Jul 13 '18 at 07:35

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You don't need tridiagonal. Gerschgorin's theorem (plus the fact the eigenvalues of a symmetric matrix are real) implies that all eigenvalues of a strictly diagonally dominant symmetric matrix with positive diagonal elements are positive, and all eigenvalues of a diagonally dominant symmetric matrix with positive diagonal elements are nonnegative.

You do need that "strictly", e.g. $\pmatrix{1 & 1\cr 1 & 1\cr}$ is diagonally dominant but not strictly diagonally dominant, and has an eigenvalue $0$.

Robert Israel
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