generalized Bézout theorem on matrices
Generalized Bézout theorem 1.
Consider given by
The polynomial can also be written as
We are now substituting the scalar argument (real or complex) by the matrix and therefore (1) and (2) will, in general, be distinct, as the powers of need not be permutable with the polynomial matrix coefficients. So that,
calling () the right (left) value of on substitution of for .
If we divide by the binomial ( is the correspondent identity matrix), we shall prove that the right (left) remainder () does not depend on . In fact,
whence we have found that
and analogously that
which proves the theorem. ∎
From this theorem we have the following
A polynomial is divisible by the characteristic polynomial on the right (left) without remainder iff ().
|Title||generalized Bézout theorem on matrices|
|Date of creation||2013-03-22 17:43:35|
|Last modified on||2013-03-22 17:43:35|
|Last modified by||perucho (2192)|