Levy-Desplanques theorem


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

|ai⁢i|>∑j≠i|ai⁢j|  ∀i;

then det⁡(A)≠0.

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

aM⁢1⁢x1+aM⁢2⁢x2+⋯+aM⁢M⁢xM+⋯+aM⁢n⁢xn=0

which implies:

|aM⁢M|⁢|xM|=|aM⁢M⁢xM|=|∑j≠MaM⁢j⁢xj|≤∑j≠M|aM⁢j|⁢|xj|≤|xM|⁢∑j≠M|aM⁢j|

that is

|aM⁢M|≤∑j≠M|aM⁢j|,

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-λ⁢I, which is singularPlanetmathPlanetmath by definition of eigenvalue: an index i must exist for which |ai⁢i-λ|=|bi⁢i|≤∑j≠in|bi⁢j|=∑j≠in|ai⁢j|, 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 |ai⁢i| is strictly greater than the same disc radius ∑j≠in|ai⁢j|, 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