Proof: Let ; then a non-zero vector exists such that ; let be the index such that , so that ; we have
Remark: the Levy-Desplanques theorem is equivalent to the well-known Gerschgorin circle theorem. In fact, let’s assume Levy-Desplanques theorem is true, and let a complex-valued matrix, with an eigenvalue ; let’s 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, let’s 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 can’t belong to any circle, so it doesn’t belong to the spectrum of , which therefore can’t be singular.
|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)|
|Author||Andrea Ambrosio (7332)|