I try to find a good proof for invertibility of strictly diagonally dominant matrices (defined by $|m_{ii}|>\sum_{j\ne i}|m_{ij}|$).
There is a proof of this in this paper but I'm wondering whether there are are better proof such as using determinant, etc to show that the matrix is non singular.
[Math] Strictly diagonally dominant matrices are non singular
linear algebramatrices
Best Answer
The proof in the PDF (Theorem 1.1) is very elementary. The crux of the argument is that if $M$ is strictly diagonally dominant and singular, then there exists a vector $u \neq 0$ with $$Mu = 0.$$
$u$ 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.
I'm skeptical you will find a significantly more elementary proof. Incidentally, though, the Gershgorin circle theorem (also described in your PDF) is very beautiful and gives geometric intuition for why no eigenvalue can be zero.