If we have a Laplacian matrix $\boldsymbol{A}$ such that 
\begin{align}
&A\_{ii} >0 \\
&A\_{ii}=-\sum\_{j\neq i}A\_{ij}
\end{align}
with known eigenvalues $\lambda\_i$.


Define the matrix $\boldsymbol{B} = \boldsymbol{A}+\boldsymbol{A}^\top$. Is there some criteria for entries of $\boldsymbol{A}$ to ensure that $\boldsymbol{B}$ is (semi-)positive definite? 



eigenvalue of Laplacian matrix
Let $z$ be the vector with all entries equal to 1. If $Az=0$ and $B=A+A^T$, then
$z^TBz=0$. Choose an orthogonal basis for $\mathbb{R}^n$ with $z$ as its first vector. If $C$ is the matrix representing $B$ relative to this basis, then $C\_{1,1}=0$. Also $B$ is positive semidefiniteif and only if $C$ is.


If $C$ is positive semidefinite and $C\_{1,1}=0$, all entries in the first row and column of $C$ must be zero. Let $C\_1$ be the matrix we get by deleting the first row and column of $C$. Then $C$ is positive semidefinite if and only if the entries in its first row and column are zero and $C\_1$ is positive semidefinite.


So we can reduce the question of whether an $n\times n$ matrix of the form $A+A^T$ is positive semidefinite to deciding whether an $(n-1)\times(n-1)$ matrix is positive semidefinite.


If $B$ is positive semidefinite and $z^TBz=0$, then $Bz=0$ and also $A^Tz=0$.
So in terms involving only $B$, we can say that $B$ is positive semidefinite
if and only if $Bz=0$ and the matrix representing the action of $B$ on the orthogonal complement to the span of $z$ is positive semidefinite.


The brief summary seems to be that knowing that $B=A+A^T$ with $Az=0$ does not give you much.


