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.