# Positive semidefiniteness of symmetric matrix with diagonal = 1 and non-diagonal elements less than 1

Does anyone know any useful results with respect to symmetric matrices with constant diagonals (specifically with respect to whether all eigenvalues are greater than $$0$$)? I am working on a set of general $$n \times n$$ matrices with diagonal entries all equal to $$1$$ and non-diagonal entries between $$0$$ and $$1$$ and I am attempting to figure out whether or not this is a subset of positive semidefinite matrices

The $$3\times3$$ matrix $$\begin{pmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{pmatrix}$$ is not p.s.d. but seems to satisfy your conditions. So the answer to your question has to depend on what particular subsets of matrices you actually have in mind.

Neither is the matrix $$\begin{pmatrix}1&1&1/2\\1&1&1\\1/2&1&1\end{pmatrix}$$.

Most $$n\times n$$ symmetric matrices are not positive definite (after all, you are asking for all $$n$$ eigenvalues to be positive). You should hook your set of matrices to one of the classical sources of psd-ness of symmetric matrices, for instance by exhibiting them as Gram matrices, or by connecting them via the Fourier Transform and Bochner's theorem to a positive measure.

• What if I restrict this further to matrices with only positive entries? – Cataclysm Nov 25 '18 at 7:15

In a symmetric matrix with positive entries, if the sum of off-diagonal entries in each row (column) is strictly less than the diagonal entry in this row (column), the matrix is called diagonally dominant and is strictly positive. See also Gershgorin theorem.

If the matrix is not diagonally dominant, the eigenvalue with the greatest absolute value (this is spectral radius of the matrix) is strictly positive, and a corresponding eigenvector with strictly positive coordinates can be choosen - see Perron-Frobenius theorem.

• Perfect thanks. This gives me good material to work from. – Cataclysm Nov 25 '18 at 9:21