polynomial functional calculus
to make sense as an element of , for any polynomial .
This is achieved in the following natural way: for any polynomial we the element .
Recall that the set of polynomial functions in , denoted by , is an associative algebra over under pointwise operations and is generated by the constant polynomial and the variable (corresponding to the identity function in ).
Definition - Consider the algebra homomorphism such that and . This homomorphism is denoted by
and it is called the polynomial functional calculus for .
It is clear that for any polynomial we have .
2 Spectral Properties
We will denote by the spectrum (http://planetmath.org/Spectrum) of an element .
Theorem - (polynomial spectral mapping theorem) - Let be an unital associative algebra over and an element in . For any polynomial we have that
: Let us first prove that . Suppose , which means that is not invertible. Now consider the polynomial in given by . It is clear that , and therefore is not invertible. Since is algebraically closed (http://planetmath.org/FundamentalTheoremOfAlgebra), we have that
Now, since is not invertible we must have that at least one of the factors is not invertible, which means that for that particular we have . But we also have that , i.e. , and hence .
We now prove the inclusion . Suppose , which means that for some . The polynomial has a zero at , hence there is a polynomial such that
Thus, we can also write a similar product for as
If was invertible, then we would see that had a left (http://planetmath.org/InversesInRings) and a right inverse (http://planetmath.org/InversesInRings), thus being invertible. But we know that , hence we conclude that cannot be invertible, i.e. .
|Title||polynomial functional calculus|
|Date of creation||2013-03-22 18:48:23|
|Last modified on||2013-03-22 18:48:23|
|Last modified by||asteroid (17536)|
|Defines||polynomial spectral mapping theorem|