# proof of determinant lower bound of a strict diagonally dominant matrix

Let’s define, for any $i=1,2,...,n$

 $h_{i}=\left|a_{ii}\right|-\sum_{j=1,j\neq i}\left|a_{ij}\right|$

Then, by strict diagonally dominance, one has $h_{i}>0$  $\forall i$. Let $D=diag\{\left(h_{1}\right)^{-1},\left(h_{2}\right)^{-1},...,\left(h_{n}\right)% ^{-1}\}$ 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

 $\displaystyle d_{i}$ $\displaystyle=$ $\displaystyle\left|b_{ii}\right|-\sum_{j=1,j\neq i}\left|b_{ij}\right|$ $\displaystyle=$ $\displaystyle\frac{\left|a_{ii}\right|}{h_{i}}-\sum_{j=1,j\neq i}\frac{\left|a% _{ij}\right|}{h_{i}}$ $\displaystyle=$ $\displaystyle 1$

Now, let $\lambda$ be an eigenvalue of $B$, and $v=[v_{1},v_{2},...,v_{n}]$ the corresponding eigenvector; let moreover $p$ be the index of the maximal component of $v$, i.e.

 $\left|v_{p}\right|\geq\left|v_{i}\right|\text{ \ }\forall i$

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

 $\displaystyle\lambda v_{p}$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{n}b_{pj}v_{j}$ $\displaystyle=$ $\displaystyle b_{pp}v_{p}+\sum_{j=1,j\neq p}^{n}b_{pj}v_{j}$

so that, being $\left|\frac{v_{j}}{v_{p}}\right|\leq 1$,

 $\displaystyle\lambda$ $\displaystyle=$ $\displaystyle b_{pp}+\sum_{j=1,j\neq p}^{n}b_{pj}\frac{v_{j}}{v_{p}}$ $\displaystyle\left|\lambda\right|$ $\displaystyle=$ $\displaystyle\left|b_{pp}+\sum_{j=1,j\neq p}^{n}b_{pj}\frac{v_{j}}{v_{p}}\right|$ $\displaystyle\geq$ $\displaystyle\left|\left|b_{pp}\right|-\left|\sum_{j=1,j\neq p}^{n}b_{pj}\frac% {v_{j}}{v_{p}}\right|\right|$ $\displaystyle\geq$ $\displaystyle\left|\left|b_{pp}\right|-\sum_{j=1,j\neq p}^{n}\left|b_{pj}% \right|\left|\frac{v_{j}}{v_{p}}\right|\right|\text{ \ \ \ \ \ \ \ \ \ (*)}$ $\displaystyle\geq$ $\displaystyle\left|\left|b_{pp}\right|-\sum_{j=1,j\neq p}^{n}\left|b_{pj}% \right|\right|\text{ \ \ \ \ \ \ \ \ \ (**)}$ $\displaystyle=$ $\displaystyle\left|b_{pp}\right|-\sum_{j=1,j\neq p}^{n}\left|b_{pj}\right|$ $\displaystyle=$ $\displaystyle d_{p}=1$

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

 $\left|\det(B)\right|=\left|\prod_{i=1}^{n}\lambda_{i}\right|\geq 1$

Finally,

 $\det(D)=\prod_{i=1}^{n}\left(h_{i}\right)^{-1}=\left(\prod_{i=1}^{n}h_{i}% \right)^{-1}$

so that

 $\displaystyle 1$ $\displaystyle\leq$ $\displaystyle\left|\det(B)\right|$ $\displaystyle=$ $\displaystyle\left|\det(D)\right|\left|\det(A)\right|$ $\displaystyle=$ $\displaystyle\left(\prod_{i=1}^{n}h_{i}\right)^{-1}\left|\det(A)\right|$

whence the thesis.

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

Title proof of determinant lower bound of a strict diagonally dominant matrix ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix 2013-03-22 17:01:11 2013-03-22 17:01:11 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 13 Andrea Ambrosio (7332) Proof msc 15-00