Levy-Desplanques theorem

A strictly diagonally dominant matrix is non-singular. In other words, let A𝐂n,n be a matrix satisfying the property

|aii|>ji|aij|  i;

then det(A)0.

Proof: Let det(A)=0; then a non-zero vector 𝐱 exists such that A𝐱=𝟎; let M be the index such that |xM|=max(|x1|,|x2|,,|xn|), so that |xj||xM|j; we have


which implies:


that is


in contrast with strictly diagonally dominance definition.

Remark: the Levy-Desplanques theorem is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to the well-known Gerschgorin circle theoremMathworldPlanetmath. In fact, let’s assume Levy-Desplanques theorem is true, and let A a n×n complex-valued matrix, with an eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath λ; let’s apply Levy-Desplanques theorem to the matrix B=A-λI, which is singularPlanetmathPlanetmath by definition of eigenvalue: an index i must exist for which |aii-λ|=|bii|jin|bij|=jin|aij|, which is Gerschgorin circle theorem. On the other hand, let’s assume Gerschgorin circle theorem is true, and let A be a strictly diagonally dominant n×n complex matrix. Then, since the absolute valueMathworldPlanetmathPlanetmath of each disc center |aii| is strictly greater than the same disc radius jin|aij|, the point λ=0 can’t belong to any circle, so it doesn’t belong to the spectrum of A, which therefore can’t be singular.

Title Levy-Desplanques theorem
Canonical name LevyDesplanquesTheorem
Date of creation 2013-03-22 15:34:50
Last modified on 2013-03-22 15:34:50
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 9
Author Andrea Ambrosio (7332)
Entry type TheoremMathworldPlanetmath
Classification msc 15-00