Page 249 - DMTH502_LINEAR_ALGEBRA

Unit 23: Semi-simple Operators

Proof: Let W be a subspace of V which is invariant under T. We must prove that W has a Notes
complementary T-invariant subspace. According to corollary of Theorem 1 of unit 20 it will
suffice to prove that if f is a polynomial and is a vector in V such that f(T) is in W, then there
is a vector in W with f(T) = f(T) . So suppose is in V and f is a polynomial such that f(T) is
in W. If f(T) = 0, we let = 0 and then is a vector in W with f(T) = f(T) . If f(T) 0, the
polynomial f is not divisible by the minimal polynomial p of the operator T. Since p is prime,
this means that f and p are relatively prime, and there exist polynomials g and h such that fg + ph
= l. Because p(T) = 0, we then have f(T)g(T) = I. From this it follows that the vector must itself be
in the subspace W; for
= g(T)f(T)
= g(T)(f(T) )
while f(T) is in W and W is invariant under T. Take = .

Theorem 1: Let T be a linear operator on the finite-dimensional vector space V. A necessary and
sufficient condition that T be semi-simple is that the minimal polynomial p for T be of the form
p = p ... p , where p , ... , p are distinct irreducible polynomials over the scalar field F.
1 k I k
Proof: Suppose T is semi-simple. We shall show that no irreducible polynomial is repeated in
the prime factorization of the minimal polynomial p. Suppose the contrary. Then there is some
2
non-scalar monic polynomial g such that g divides p. Let W be the null space of the operator g(T).
2
Then W is invariant under T. Now p = g h for some polynomial h. Since g is not a scalar polynomial,
the operator g(T)h(T) is not the zero operator, and there is some vector in V such that g(T)h(T)
0, i.e., (gh) 0. Now (gh) is in the subspace W, since g(gh ) = g h = p = 0. But there is no
2
vector in W such that gh = qh ; for, if is ill W
(gh) = (hg) = h(g ) = h(0) = 0.
Thus, W cannot have a complementary T-invariant subspace, contradicting the hypothesis that T
is semi-simple.
Now suppose the prime factorization of p is p = p ... p , where p , ... , p are distinct irreducible
1 k 1 k
(non-scalar) monic polynomials. Let W be a subspace of V which is invariant under T. We shall
prove that W has a complementary T-invariant subspace. Let V = W ... W be the primary
I k
decomposition for T, i.e., let W be the null space of p (T). Let T be the linear operator induced on
j j j
W by T, so that the minimal polynomial for T is the prime p . Now W W is a subspace of W
j j j j j
which is invariant under T (or under T). By the last lemma, there is a subspace V of W such that
j j j
W = (W W ) V and V is invariant under T (and hence under T). Then we have
j j j j j
V = W ... W
1 k
= (W W ) V ... (W W ) V
1 1 k k
= (W W ) + ... + (W W ) V ... V .
1 k 1 k
By the first lemma above, W = (W W ) ... (W W ) so that if W’ = V ... V , then V = W
1 k 1 k
W’ and W’ is invariant under T.
Corollary: If T is a linear operator on a finite-dimensional vector space over an algebraically
closed field, then T is semi-simple if and only if T is diagonalizable.

Proof: If the scalar field F is algebraically closed, the monic primes over F are the polynomials
x – c. In this case, T is semi-simple if and only if the minimal polynomial for T is p = (x – c ) ...
1
(x – c ), where c , ... , c are distinct elements of F. This is precisely the criterion for T to be
k 1 k
diagonalizable.
We turn now to expressing a linear operator as the sum of a semi-simple operator and a nilpotent
operator which commute. In this, we shall restrict the scalar field to a subfield of the complex

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