proof of determinant lower bound of a strict diagonally dominant matrix

Let’s define, for any $i=1,2,\mathrm{\dots },n$

$$h_i=\left|a_{ii}\right|-\sum _{j=1,j\ne i}\left|a_{ij}\right|$$

Then, by strict diagonally dominance, one has $h_i>0$  $\forall i$.Let $D=diag\{{\left(h_1\right)}^{-1},{\left(h_2\right)}^{-1},\mathrm{\dots },{\left(h_n\right)}^{-1}\}$ and $B=DA$, so that the i-th row of $B$ matrix is equal to the corresponding row of $A$ matrix multiplied by ${\left(h_i\right)}^{-1}$. In this way , one has

$d_i$$=$$\left|b_{ii}\right|-{\displaystyle \sum _{j=1,j\ne i}}\left|b_{ij}\right|$

$=$${\displaystyle \frac{\left|a_{ii}\right|}{h_i}}-{\displaystyle \sum _{j=1,j\ne i}}{\displaystyle \frac{\left|a_{ij}\right|}{h_i}}$

$=$$1$

Now, let $\lambda$ be an eigenvalue of $B$, and $v=[v_1,v_2,\mathrm{\dots },v_n]$the corresponding eigenvector; let moreover $p$ be the index of the maximalcomponent of $v$, i.e.

$$\left|v_p\right|\ge \left|v_i\right|\forall i$$

Of course, by definition of eigenvector, $\left|v_p\right|>0$.Writing the p-th characteristic equation, we have:

$\lambda v_p$$=$${\displaystyle \sum _{j=1}^n}{b}_{pj}{v}_j$

$=$$b_{pp}{v}_p+{\displaystyle \sum _{j=1,j\ne p}^n}{b}_{pj}{v}_j$

so that, being $\left|\frac{v_j}{v_p}\right|\le 1$,

$\lambda$$=$$b_{pp}+{\displaystyle \sum _{j=1,j\ne p}^n}{b}_{pj}{\displaystyle \frac{v_j}{v_p}}$

$\left|\lambda \right|$$=$$\left|b_{pp}+{\displaystyle \sum _{j=1,j\ne p}^n}{b}_{pj}{\displaystyle \frac{v_j}{v_p}}\right|$

$\ge$$\left|\left|b_{pp}\right|-\left|{\displaystyle \sum _{j=1,j\ne p}^n}{b}_{pj}{\displaystyle \frac{v_j}{v_p}}\right|\right|$

$\ge$$\left|\left|b_{pp}\right|-{\displaystyle \sum _{j=1,j\ne p}^n}\left|b_{pj}\right|\left|{\displaystyle \frac{v_j}{v_p}}\right|\right|\text{ (*)}$

$\ge$$||b_{pp}\left|-{\displaystyle \sum _{j=1,j\ne p}^n}\right|b_{pj}||\text{ (**)}$

$=$$\left|b_{pp}\right|-{\displaystyle \sum _{j=1,j\ne p}^n}\left|b_{pj}\right|$

$=$$d_p=1$

In this way, we found that each eigenvalue of $B$ is greater than one inabsolute value; for this reason,

$$\left|det(B)\right|=\left|\prod _{i=1}^n{\lambda }_i\right|\ge 1$$

Finally,

$$det(D)=\prod _{i=1}^n{\left(h_i\right)}^{-1}={\left(\prod _{i=1}^n{h}_i\right)}^{-1}$$

so that

$1$$\le$$\left|det(B)\right|$

$=$$\left|det(D)\right|\left|det(A)\right|$

$=$${\left({\displaystyle \prod _{i=1}^n}{h}_i\right)}^{-1}\left|det(A)\right|$

whence the thesis.

Remark: Perhaps it could be not immediately evident where thehypothesis of strict diagonally dominance is employed in this proof; infact, inequality (*) and (**) would be, in a general case, not valid; they can be stated only because we can assure, by virtue of strictdiagonally dominance, that the final argument of the absolute value ($\left|b_{pp}\right|-{\sum }_{j=1,j\ne p}^n\left|b_{pj}\right|$) does remain positive.