proof of determinant lower bound of a strict diagonally dominant matrix

Let’s define, for any i=1,2,,n


Then, by strict diagonally dominance, one has hi>0  i. Let D=diag{(h1)-1,(h2)-1,,(hn)-1} and B=DA, so that the i-th row of B matrix is equal to the corresponding row of A matrix multiplied by (hi)-1. In this way , one has

di = |bii|-j=1,ji|bij|
= |aii|hi-j=1,ji|aij|hi
= 1

Now, let λ be an eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath of B, and v=[v1,v2,,vn] the corresponding eigenvectorMathworldPlanetmathPlanetmathPlanetmath; let moreover p be the index of the maximal componentPlanetmathPlanetmathPlanetmath of v, i.e.

|vp||vi| i

Of course, by definition of eigenvector, |vp|>0. Writing the p-th characteristic equationMathworldPlanetmathPlanetmath, we have:

λvp = j=1nbpjvj
= bppvp+j=1,jpnbpjvj

so that, being |vjvp|1,

λ = bpp+j=1,jpnbpjvjvp
|λ| = |bpp+j=1,jpnbpjvjvp|
||bpp|-j=1,jpn|bpj||vjvp|| (*)
||bpp|-j=1,jpn|bpj|| (**)
= |bpp|-j=1,jpn|bpj|
= dp=1

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




so that

1 |det(B)|
= |det(D)||det(A)|
= (i=1nhi)-1|det(A)|

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 (|bpp|-j=1,jpn|bpj|) does remain positive.

Title proof of determinantMathworldPlanetmath lower bound of a strict diagonally dominant matrixMathworldPlanetmath
Canonical name ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix
Date of creation 2013-03-22 17:01:11
Last modified on 2013-03-22 17:01:11
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 13
Author Andrea Ambrosio (7332)
Entry type Proof
Classification msc 15-00