0

I am working on solving a system of linear equations which ended with a diagonally dominant matrix whose main diagonal is strictly positive ($>0$), while off-diagonal entries are nonpositive ($\leq 0$), it is not strictly diagonally dominant, but there are some rows satisfy the strict dominance. I also proved that the matrix is irreducible. Does that guarantee the invertibility of the matrix?

I am looking for some relationships between these families of matrices (reducible, "strictly" diagonally dominant, ..) and try to find any interesting characterizations for them. If there is a textbook you recommend me to read that address these concepts with some examples and introduce characterizations, I would really appreciate that.

Adam_math
  • 329
  • 1
  • 8
  • 1
    Yes. An irreducible diagonally dominant matrix with strict diagonal dominance occurring in at least one row is called irreducibly diagonally dominant. Olga Taussky was the first to give a rigorous proof that such a matrix is invertible. See also my answer under another question. – user1551 Feb 06 '25 at 17:08
  • 1
    I gave a proof of Taussky's Gerschgorin discs refinement here https://math.stackexchange.com/questions/3903369/prove-that-this-block-matrix-is-positive-definite/3904956#3904956 – user8675309 Feb 06 '25 at 17:15
  • @user1551, Can we deduce that this matrix with these properties is monotone? – Adam_math Feb 08 '25 at 14:56
  • Yes. By permuting the rows and columns simultaneously, we may assume that strict diagonal dominance occurs on the first row. All leading principal minors of the matrix are therefore positive. By mathematical induction and by LU decomposition, one may prove that the matrix is monotone (i.e., inverse-positive). The authoritative reference for this kind of stuffs is Nonnegative Matrices in the Mathematical Sciences by Berman and Plemmons. See also the last paragraph of my answer under another question. – user1551 Feb 08 '25 at 15:24
  • I think it is also possible to prove inverse-positivity using Neumann series, but care must be taken because the matrix is not strictly diagonally dominant. – user1551 Feb 08 '25 at 15:25

0 Answers0