Is a symmetric real matrix with diagonal entries strictly greater than $1$ and off-diagonal entries positive but strictly less than $1$ necessarily positive-semidefinite?
Positive symmetric matrices and positive-definiteness
3 Answers
Nope. Just playing around with my computer, I found the matrix
[$\frac{11}{10}$ $\frac{1}{100}$ $\frac{99}{100}$]
[$\frac{1}{100}$ $\frac{11}{10}$ $\frac{99}{100}$]
[$\frac{99}{100}$ $\frac{99}{100}$ $\frac{11}{10}$]
with determinant $\frac{-25179}{31250}$.
Is this perhaps a misremembering of the definition of a diagonally dominant matrix?
-
0@daegan: no problem. To construct a counterexample, it of course helps to be aware of the theorem about diagonally dominant matrices, so as to stay away from that case. I did remember that theorem (luckily for me, I learned about Gershgorin's Theorem in a talk I heard last year) so the above was only the third matrix I tried (or technically the fourth: the third one also worked but the coefficients were a little more complicated: it appears in a previous version of this answer). – 2011-02-08
If the sum of each row is positive, then no eigenvalue can be non-negative: take any eigenvector, and pick the largest (in magnitude) coordinate. After applying the matrix, even if the signs are all perfectly against us, still $0$ will not be crossed (or reached to).
So the correct formulation is $\geq 1$ on the diagonal, and the off-diagonals in each row sum to less than $1$.
By Gershgorin's Theorem we can conclude that the absolute sum of all non-diagonal entries in each row must be less than 1. Then the matrix will have positive eigenvalues for diagonal entries > 1.