How to prove that a matrix is positive definite?

Question

Let $L$ be a Laplacian matrix of a strong connected and balanced directed graph. Define $$ L^{s}=\frac{1}{2}\left( L+L^{T}\right) .$$ Let $D$ be a diagonal matrix with $$ D=\begin{bmatrix} d_{1} & & & \\ & d_{2} & & \\ & & \ddots & \\ & & & d_{n}% \end{bmatrix}, $$ with $d_{i}\geq 0.$ There is at least one $d_{i}>0$. Clearly, this matrix is positive semi-definite. Is the matrix $ L^{s}+D $ positive definite or not?

Answer 1

Take any nonzero $\mathbf{x}\in\mathbf{R}^n$ and let $\mathbf{x}=(x_1,x_2,\dotsc ,x_n)$.

Then

$\quad \mathbf{x} L_s\mathbf{x}^t=\sum_{ij} (x_i-x_j)^2 > 0$,

where the sum is taken over the edges of the graph (edges without orientation), which shows that $L_s$ is positive definite (the graph is strongly connected). On the other hand $D$ is clearly positive semi-definite, and hence the addition of both matrices is positive definite.