Definition.
An $n\times n$ complex matrix $M$ is strictly diagonally dominant if
\[
|m_{ii}|>\sum_{j\neq i}|m_{ij}|\\ (i=1,...,n).
\]
Theorem.
If $M$ is strictly diagonally dominant then invertible.
Indirectly, assume there exists a vector $\mathbf{u}\neq\mathbf{0}$ such that
\[ M\mathbf{u}=\mathbf{0}. \] $\mathbf{u}$ has some entry $|u_i|>0$ of largest magnitude. Then
\[
\begin{align*} \sum_j m_{ij} u_j &= 0\\ m_{ii} u_i &= -\sum_{j\neq i} m_{ij}u_j\\ m_{ii} &= -\sum_{j\neq i} \frac{u_j}{u_i}m_{ij}\\ |m_{ii}| &\leq \sum_{j\neq i} \left|\frac{u_j}{u_i}m_{ij}\right|\\ |m_{ii}| &\leq \sum_{j\neq i} |m_{ij}|, \end{align*} \] a contradiction. $\blacksquare$
\[ M\mathbf{u}=\mathbf{0}. \] $\mathbf{u}$ has some entry $|u_i|>0$ of largest magnitude. Then
\[
\begin{align*} \sum_j m_{ij} u_j &= 0\\ m_{ii} u_i &= -\sum_{j\neq i} m_{ij}u_j\\ m_{ii} &= -\sum_{j\neq i} \frac{u_j}{u_i}m_{ij}\\ |m_{ii}| &\leq \sum_{j\neq i} \left|\frac{u_j}{u_i}m_{ij}\right|\\ |m_{ii}| &\leq \sum_{j\neq i} |m_{ij}|, \end{align*} \] a contradiction. $\blacksquare$
Corollary.
Let $n>1$ integer and $\widehat{\mathbf{v}}_1,\ldots, \widehat{\mathbf{v}}_n$ unit vectors in an inner product space. If\[
|\langle \widehat{\mathbf{v}}_j, \widehat{\mathbf{v}}_k\rangle|<\frac{1}{n-1}\qquad \forall j\neq k
\]
then $\{\widehat{\mathbf{v}}_1,\ldots,\widehat{\mathbf{v}}_n\}$ is a linearly independent system.
We can prove Corollary directly.
Indeed,
\[\sum_i\alpha_i \widehat{\mathbf v}_i=0\Leftrightarrow \left\|\sum_i\alpha_i \widehat{\mathbf v}_i\right\|^2=\sum_i|\alpha_i|^2+\sum_{i\neq j}\alpha_i\overline {\alpha }_{j}\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle=0,\]
that is,
\[\sum_i|\alpha_i|^2=\left|\sum_{i\neq j}\alpha_i\overline {\alpha }_{j}\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle\right|\le
\sum_{i\neq j}|\alpha_i||\overline {\alpha }_{j}||\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle|\stackrel{\exists \alpha_i\alpha_j\ne 0}<\frac1{n-1}
\sum_{i\neq j}|\alpha_i||\alpha_j| \]
\[
\le\frac1{n-1}
\sum_{i\neq j}\frac{|\alpha_i|^2+|\alpha_j|^2}2=\sum_i|\alpha_i|^2,
\]
a contradiction. $\blacksquare$
Indeed,
\[\sum_i\alpha_i \widehat{\mathbf v}_i=0\Leftrightarrow \left\|\sum_i\alpha_i \widehat{\mathbf v}_i\right\|^2=\sum_i|\alpha_i|^2+\sum_{i\neq j}\alpha_i\overline {\alpha }_{j}\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle=0,\]
that is,
\[\sum_i|\alpha_i|^2=\left|\sum_{i\neq j}\alpha_i\overline {\alpha }_{j}\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle\right|\le
\sum_{i\neq j}|\alpha_i||\overline {\alpha }_{j}||\langle\widehat{\mathbf v}_i,\widehat{\mathbf v}_j\rangle|\stackrel{\exists \alpha_i\alpha_j\ne 0}<\frac1{n-1}
\sum_{i\neq j}|\alpha_i||\alpha_j| \]
\[
\le\frac1{n-1}
\sum_{i\neq j}\frac{|\alpha_i|^2+|\alpha_j|^2}2=\sum_i|\alpha_i|^2,
\]
a contradiction. $\blacksquare$
