Positive semi-definition of Diagonal Dominant Matrix (with non-negative diagonal) is an important property in numeric linear algebra. Here I record two prooves, one following the idea of Gershgorin Circle Theorem, the other is a proof for the weakened version of strictly diagonal dominance.
A diagonal dominant matrix \(A=\{a_{ij}\}_{i,j=1}^n\) satisties \(\begin{align} \vert a_{ii}\vert \geq \sum_{j\neq i}\vert a_{ij}\vert ,\quad \forall i \end{align}\)
Gershgorin Circle Theorem
The eigen vector \(x\) with eigen value \(\lambda\), say \(x_i\) is the element with largest absolute value. \(\begin{align} \sum_{j=1}^na_{ij}x_{j}=\lambda x_i\Rightarrow \vert (\lambda -a_{ii})\vert =&\left\vert \sum_{j\neq i}\dfrac{a_{ij}x_j}{x_i}\right\vert \leq \sum_{j\neq i} \left\vert a_{ij}\right\vert \left\vert \dfrac{x_{j}}{x_i}\right\vert \\ \leq& \sum_{j\neq i}\vert a_{ij}\vert \end{align}\)
i.e. \(\lambda\) lies in one of the Gershgorin disks: \(\begin{align} \lambda \in \Cup_{i=1}^n \mathrm{Disc}\left( a_{ii}; R_i=\sum_{j\neq i}\vert a_{ij}\vert \right) \end{align}\)
Immediately we would find that for diagonal dominant matrix with non-negative diagonal elements, all eigen vectors would be non-negative, thus \(A\) is positive semi-definite. \(\begin{align} a_{ii}-\lambda \leq \vert \lambda -a_{ii}\vert \leq \sum_{j\neq i}\vert a_{ij}\vert \Rightarrow \ \lambda \geq \vert a_{ii}\vert -\sum_{j\neq i}\vert a_{ij}\vert \geq 0 \end{align}\)
Another Interesting Proof for Strictly Diagonal Dominant
Strictly Diagonal Dominant Matrix: \(\begin{align} \vert a_{ii}\vert >\sum_{j\neq i}\vert a_{ij}\vert ,\quad \forall i \end{align}\)
A strictly diagonal dominant matrix is non-singular: Assume there \(\exists x, s.t. Ax=0\), with \(x_i\) the element with largest absolute value, then follows similar idea as in Gershgorin thm.: \(\begin{align} \sum_{j=1}^na_{ij}x_j=0\Rightarrow \vert a_{ii}\vert =\left\vert \sum_{j\neq i} \dfrac{a_{ij}x_j}{x_i} \right\vert \leq \sum_{j\neq i}\vert a_{ij}\vert \end{align}\)
which contradicts with strict diagonal dominance, thus \(A\) is non-singular. i.e. \(\vert A\vert \neq 0\)
Further consider matrix \(A+\tau I\), which is also strictly diagonal dominant, thus \(\vert A+\tau I\vert \neq 0\). Consider the function \(\begin{align} \phi (\tau):=\vert A+\tau I\vert \neq 0,\quad \forall \tau\geq 0 \end{align}\)
Naturally \(\phi (\tau)\) should be continuous, and \(\lim_{\tau\to\infty}\phi(\tau)>0\), which would indicate that \(\begin{align} \phi (0)=\vert A\vert >0 \end{align}\)
Notice that all the sequential principal minor \(D_i\)s of \(A\) are still strictly diagonal dominant, thus \(\vert D_i\vert >0\), \(\forall i=1,2,\ldots,n\). Thus \(A\) is positive definite.