Is this kind of matrix positive semi-definite?

Consider a real symmetric matrix $A$ with positive diagonal elements and zero row sums. i.e.
$$A_{ii}>0,\;\sum_jA_{ij}=0,\;\forall i,j$$ $A$ is not necessarily positive semi-definite, for example, let $A=\begin{bmatrix}1&2&-3\\2&1&-3\\-3&-3&6\end{bmatrix}$ and $A$ has an eigenvalue -1.

Here is my question: Suppose we also know that each diagonal element of $A$ is larger than the absolute value of any other element in the same row, i.e.
$$A_{ii}\geq|A_{ij}|,\;\forall i\neq j.$$

Is the matrix $A$ positive semi-definite?

I have tried several cases using Matlab, and the results show it is positive semi-definite. But I cannot prove. Is there any suggestion?

Thanks!

EDIT: I know if the off-digonal elements of $A$ are all non-positive then $A$ is positive semi-definite, which could be proved by the Gershgorin circle theorem. But in my question the off-digonal elements may be positive.

• The Gershgorin circle theorem implies that if $A_{ii} \ge \sum_{j \ne i} |A_{ij}|$ for each $i$ (diagonally dominant matrix), then the matrix is PSD. This is a stronger condition than your condition, however. – angryavian Jul 9 '17 at 7:28
• @angryavian I just assume $A_{ii}\geq|A_{ij}|$, which can not conclude $A_{ii}\geq\sum_{j\neq i}|A_{ij}|$. In fact, $A_{ij}$ may be positive. So the Gershgorin circle theorem doesn't work here. – Jiaqi Jul 9 '17 at 7:39

Try $$\pmatrix{ 1 & 1 & -1 & 0 & -1\cr 1 & 1 & 0 & -1 & -1\cr -1 & 0 & 1 & 1 & -1\cr 0 & -1 & 1 & 1 & -1\cr -1 & -1 & -1 & -1 & 4\cr }$$