eigenvalue of Laplacian matrix

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?

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If $A$ is symmetric (i.e., the corresponding graph is undirected), then $B$ will always be positive semidefinite. This is true because every symmetric Laplacian matrix is positive semidefinite. And being symmetric it shares eigenvalues with its transpose. Specifically,

$$\lambda(A)_0 = \lambda(A^T)_0 \geq 0.$$

So the smallest eigenvalue of $B$ is given by

$$\lambda(B)_0 =\arg \min_x x^T(A+A^T)x=\arg \min_x x^TAx+x^TA^Tx = 2\lambda(A)_0 \geq 0$$

It follows that $B$ is positive semidefinite since it's eigenvalues are all non-negative.

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Also: if $A$ is symmetric, $A=A^T$ and $B=2A$. –  Federico Poloni Nov 10 at 9:08

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.

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Thank you very much –  user98883 Oct 14 at 16:36
You want the numerical range, or field of values, of $A$ to be included in the right half-plane.