Page 251 - DMTH502_LINEAR_ALGEBRA

Unit 23: Semi-simple Operators

n Notes
f' = p f
j j
j 1
we see that p must divide p’f. Therefore p divides either f or p’. But p does not divide f , since p ,
i i i i i i i i 1
... , p are distinct. So p divides p'. This is not possible, since p’ has degree one less than the degree
k i i i
of p . We conclude that no prime divides both f and f', or that (f, f') = 1.
i
To see that statement (c) is equivalent to (a) and (b), we need only observe the following:
Suppose f and g are polynomials over F, a subfield of the complex numbers. We may also regard
f and g as polynomials with complex coefficients. The statement that f and g are relatively prime
as polynomials over F is equivalent to the statement that f and g are relatively prime as
polynomials over the field of complex numbers. We use this fact with g = f'. Note that (c) is just
(a) when f is regarded as a polynomial over the field of complex numbers. Thus (b) and (c) are
equivalent, by the same argument that we used above.
Theorem 2: Let F be a subfield of the field of complex numbers, let V be a finite-dimensional
vector space over F, and let T be a linear operator on V. Let  be an ordered basis for V and let
A be the matrix of T in the ordered basis . Then T is semi-simple if and only if the matrix A is
similar over the field of complex numbers to a diagonal matrix.

Proof: Let p be the minimal polynomial for T. According to Theorem 1, T is semi-simple if and
only if p = p ... p where p , ... , p , are distinct irreducible polynomials over F. By the last lemma,
1 k 1 k
we see that T is semi-simple if and only if p has no repeated complex root.
Now p is also the minimal polynomial for the matrix A. We know that A is similar over the field
of complex numbers to a diagonal matrix if and only if its minimal polynomial has no repeated
complex root. This proves the theorem.
Theorem 3: Let F be a subfield of the field of complex numbers, let V be a finite-dimensional
vector space over F, and let T be a linear operator on V. There is a semi-simple operator S on V
and a nilpotent operator N on V such that
(i) T = S + N;

(ii) SN = NS.
Furthermore, the semi-simple S and nilpotent N satisfying (i) and (ii) are unique, and each is a
polynomial in T.
k r
p 1 r ... p be the prime factorization of the minimal polynomial for T, and let f = p ...
Proof: Let 1 k 1
p · Let r be the greatest of the positive integers r , ... , r , Then the polynomial f is a product of
k 1 k
distinct primes, f' is divisible by the minimal polynomial for T, and so
r
f(T) = 0.
We are going to construct a sequence of polynomials: g , g , g , ... such that
0 1 2
n
f x g f j
j
j 0
0
n+1
is divisible by f , n = 0, 1,2, .... We take g = 0 and then f(x – g f = f(x) = f is divisible by f. Suppose
0 0
we have chosen g , ... , g . Let
0 n–1
n 1
h = x g f j
j
j 0

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