Levy-Desplanques theorem
A strictly diagonally dominant matrix is non-singular. In other words, let Aβπn,n be a matrix satisfying the property
|aii|>βjβ i|aij|βββi; |
then .
Proof: Let ; then a non-zero vector exists such that ; let be the index such that , so that ; we have
which implies:
that is
in contrast with strictly diagonally dominance definition.
Remark:
the Levy-Desplanques theorem is equivalent to the well-known Gerschgorin circle theorem
. In fact, letβs assume Levy-Desplanques theorem is true, and let a complex-valued matrix, with an eigenvalue
; letβs apply Levy-Desplanques theorem to the matrix , which is singular
by definition of eigenvalue: an index must exist for which , which is Gerschgorin circle theorem.
On the other hand, letβs assume Gerschgorin circle theorem is true, and let be a strictly diagonally dominant complex matrix. Then, since the absolute value
of each disc center is strictly greater than the same disc radius , the point canβt belong to any circle, so it doesnβt belong to the spectrum of , which therefore canβt 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 | Theorem![]() |
Classification | msc 15-00 |