Page 250 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes numbers. We will see that what is important is that the field F be a field of characteristic zero,
that is, that for each positive integer n the sum 1 + ... + 1 (n times) in F should not be 0. For a
(k)
(k)
k
polynomial f over F, we denote by f the kth formal derivative of f. In other words, f = D f, where
D is the differentiation operator on the space of polynomials. If g is another polynomial, f(g)
denotes the result of substituting g in f, i.e., the polynomial obtained by applying f to the
element g in the linear algebra F[x].
Lemma (Taylor's Formula): Let F be a field of characteristic zero and let g and h be polynomials
over F. If f is any polynomial over F with deg f n, then
n
h
h
f (2) ( ) f ( ) ( )
f(g) = f(h) + f (h)(g – h) + (g – h) + ... + (g – h) ,
(1)
2
n
2! ! n
Proof: What we are proving is a generalized Taylor formula. The reader is probably used to
seeing the special case in which h = c, a scalar polynomial, and g = x. Then the formula says
n
c
f (2) ( ) f ( ) ( )
c
f = f(x) = f(c) + f (c) (x – c) + (x – c) + ... + (x – c) n
2
(1)
2! ! n
The proof of the general formula is just an application of the binomial theorem
( k k 1) k 2 2
(a + b) = a + ka b + a b + ... + b .
k-1
k
k
k
2!
Since substitution and differentiation are linear processes, one need only prove the formula
n
k
when f = x . The formula for f = c x follows by a linear combination. In the case f = x with
k
k
k
k 0
k n, the formula says
( k k 1)
2
k–l
k–2
k
k
g = h + kh (g – h) + h (g – h) + ... + (g – h) k
2!
which is just the binomial expansion of
g = [h + (g – h)] . k
k
Lemma: Let F be a subfield of the complex numbers, let f be a polynomial over F, and let f' be the
derivative of f. The following are equivalent:
(a) f is the product of distinct polynomials irreducible over F.
(b) f and f' are relatively prime.
(c) As a polynomial with complex coefficients, f has no repeated root.
Proof: Let us first prove that (a) and (b) are equivalent statements about f. Suppose in the prime
factorization of f over the field F that some (non-scalar) prime polynomial p is repeated. Then f
= p h for some h in F[x]. Then
2
2
f' = p h' + 2pp'h
and p is also a divisor of f'. Hence f and f' are not relatively prime. We conclude that (b) implies (a).
Now suppose f = p ... p , where p , ... , p are distinct non-scalar irreducible polynomials over F.
1 k I k
Let f = f /p . Then
j j
f' = P’f + P’ f + ... + P’ f
1 1 2 2 k k
Let p be a prime polynomial which divides both f and f '. Then p = p , for some i. Now p divides
i i
f for j i, and since p also divides
j i
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