Page 252 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes so that, by assumption, f(h) is divisible by f . We want to choose g so that f(h) is divisible by f .
n
n+1
n
We apply the general Taylor formula and obtain
n
n+l
f(h – g f ) = f(h) – g f f'(h) + f b
n
n n
n
n
where b is some polynomial. By assumption f(h) = qf . Thus, we see that to have f(h – g f ) divisible
n
by f we need only choose g in such a way that (q – g f’) is divisible by f. This can be done,
n+1
n n
because f has no repeated prime factors and so f and f’ are relatively prime. If a and e are
polynomials such that af + ef' = 1, and if we let g = eq, then q – g f’ is divisible by f.
n n
n
n+1
Now we have a sequence g , g , .. , such that f divides f x g f i . Let us take n = r – 1 and
j
0 1
j 0
then since f(T) = 0
r
r 1
f T g j ( ) ( ) i = 0.
f
T
T
j 0
Let
r 1 r 1
j j
T
T
f
f
T
T
N = g j ( ) ( ) g j ( ) ( )
j 0 j 0
n
i
Since g f is divisible by f, we see that N = 0 and N is nilpotent. Let S = T – N. Then f(S) =
r
j
j 1
f(T – N) = 0. Since f has distinct prime factors, S is semi-simple.
Now we have T = S + N where S is semi-simple, N is nilpotent, and each is a polynomial in T. To
prove the uniqueness statement, we shall pass from the scalar field F to the field of complex
numbers. Let be some ordered basis for the space V. Then we have
[T] = [S] + [N]
while [S] is diagonalizable over the complex numbers and [N] is nilpotent. This diagonalizable
matrix and nilpotent matrix which commute are uniquely determined.
Self Assessment
1. If N is a nilpotent linear operator on V, show that for any polynomial f the semi-simple
part of f(N) is a scalar multiple of the identity operator (F a subfield of C).
2. Let T be a linear operator on R which is represented by the matrix
3
3 1 1
2 2 1
2 2 0
in the standard ordered basis. Show that there is a semi-simple operator S on R and a
3
nilpotent operator N on V such that T = S + N and SN = NS.
23.3 Summary
 In this unit the idea of semi-simple linear operator is explored after a brief review of the
outcome of the previous few units.
 It is shown that a linear operator is semi-simple if every T-invariant subspace W of the
finite dimensional space V, has a complementary T-invariant subspace W’ such that
V = W W’.

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