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Levy-Desplanques theorem


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

|aii|>βˆ‘jβ‰ i|aij|β€ƒβ€ƒβˆ€i;

then det⁑(A)β‰ 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|,β‹―,|xn|), so that |xj|≀|xM|β€ƒβˆ€j; we have

aM⁒1⁒x1+aM⁒2⁒x2+β‹―+aM⁒M⁒xM+β‹―+aM⁒n⁒xn=0

which implies:

|aM⁒M|⁒|xM|=|aM⁒M⁒xM|=|βˆ‘jβ‰ MaM⁒j⁒xj|β‰€βˆ‘jβ‰ M|aM⁒j|⁒|xj|≀|xM|β’βˆ‘jβ‰ M|aM⁒j|

that is

|aM⁒M|β‰€βˆ‘jβ‰ M|aM⁒j|,

in contrast with strictly diagonally dominance definition.β–‘

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 nΓ—n complex-valued matrix, with an eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath Ξ»; 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-Ξ»|=|bi⁒i|β‰€βˆ‘jβ‰ in|bi⁒j|=βˆ‘jβ‰ 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 nΓ—n complex matrix. Then, since the absolute valueMathworldPlanetmathPlanetmath of each disc center |ai⁒i| is strictly greater than the same disc radius βˆ‘jβ‰ in|ai⁒j|, the point Ξ»=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