# Ensuring that a symmetric matrix with nonnegative elements is positive semidefinite

I have the following matrix $A$:

• symmetric
• all positive and/or zero values
• the main diagonal is all the same value, $x$.

To ensure that the matrix $A$, is positive semidefinite, must I only ensure that $x \geq 0$? It seems correct from my thinking, but wanted to make sure. Thanks.

It is not sufficient to have positive diagonal entries. To see this, consider the matrix $$A=\pmatrix{1& 10\\10& 1}.$$ It has the negative eigenvalue $-9$ to the eigenvector $$v=\pmatrix{1\\-1},$$and is thus not positive semi-definite.

Note however that a diagonally dominant symmetric matrix is positive semi definite.

In your case it is sufficient to add the condition $x \ge \sum_{j\neq i} A_{ij}, i\neq j$ to ensure $A$ to be positive semi definite.

• Why does it suffice for $x\ge A_{ij}$? I would appreciate a reference for this if you have one. – user3281410 Feb 28 '18 at 1:25
• @user3281410 The wikipedia article states this property for Hermitian diagonally dominant matrices, so it applies for real symmetric matrices as well : en.wikipedia.org/wiki/Diagonally_dominant_matrix – Tool Mar 17 '18 at 15:06
• I believe you mean to say $x\ge \sum_{j\neq i}^{}A_{ij}$ for all $i$, not just $x\ge A_{ij}$ for all $i\neq j$. There are symmetric, matrices with nonnegative entries such that the entries on the diagonal are all the same (denoted $x$) and satisfy $x \ge A_{ij}$ but $A$ is not positive semidefinite. See the first answer here: math.stackexchange.com/questions/2669941/… – user3281410 Mar 20 '18 at 22:56
• @user3281410 Yep sorry I really meant $x \ge \sum_j A_{ij}, i\neq j$. I've updated the answer ! – Tool Mar 21 '18 at 14:36
• @Tool is the converse statement true? If I have a positive semidefinite matrix, can I say that it must also be diagonally dominant with nonnegative real entries on the diagonal? – Tyberius Mar 28 '18 at 20:12