Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A common characterization of M-matrices are non-singular square matrices with non-positive off-diagonal entries, positive diagonal entries, non-negative row sums, and at least one positive row sum

It seems that this characterization depends on the use of rows or columns. I don't understand how that is possible, as the transposed of an M-Matrix is also an M-Matrix.

Example: $\left( \begin{array}{ccc} 2 & -1 &0\\ -2 & 2&-1\\ 0&-1&2 \end{array} \right)$

This matrix is not diagonally dominant, but its transpose is.

However the matrix is clearly M-Matrix, because all Eigenvalues are positive.

Thanks for your help!

share|cite|improve this question
Isn't the second row sum negative for your example? – hardmath Jan 12 '11 at 19:02
yes. its not diagonally dominant. the row sum is negative. and its still a M-Matrix. – JSG Jan 12 '11 at 19:30
I'm afraid Wikipedia has let you down here. Unless one assumes "characterization" was meant to imply a sufficient condition rather that a definition, the statement quoted is not correct. Some authors include singular as well as non-singular M-matrices, but if we restrict attention to the non-singular one, then the first part of the Wikipedia article is correct. – hardmath Jan 12 '11 at 19:41
up vote 1 down vote accepted

There are many characterizations of non-singular M-matrices which make it clear that the property is preserved by taking transposes. For example, Berman and Plemmons book, Nonnegative matrices in the mathematical sciences, devotes Chapt. 6 to M-matrices. Thm. 2.3 there gives many equivalences to a Z-matrix (that is, a matrix with nonpositive off-diagonal entries) to be a non-singular M-matrix. But the condition you've cited is not among them.

One equivalence given there ($A_1$) is for all the principal minors to be positive. Clearly this is preserved by transpose.

The sufficient condition you described is not preserved by taking transpose, as your example shows. However it is certainly an M-matrix. The cited sufficient condition is closely related to weak diagonal dominance, and it seems to be an echo of the result by Minkowski (that a Z-matrix with all positive row sums has positive determinant) which apparently motivated Ostrowski's choice of the term M-matrix (honoring Minkowski).

Added: I've corrected the Wikipedia article. Thanks for pointing out this mistake.

share|cite|improve this answer

See this and this. I think your characterization of M-matrices is a sufficient one but not necessary.

share|cite|improve this answer
THank you! I accepted the answer that was first! – JSG Jan 12 '11 at 20:28
@user4514: My answer was first. – PEV Jan 14 '11 at 6:13
@user4514: It is true, Trevor's answer was posted 11 minutes and 56 seconds before hardmath's. You can see the exact times posted by mousing over the time to the right of "answered". E.g., right now it says "answered yesterday", but mousing over "yesterday" reveals "2011-01-12 19:40:53Z". – Jonas Meyer Jan 14 '11 at 6:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.