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P. 324
Linear Algebra
Notes Then
W = W W j
j
Proof: Clearly W contains W W . On the other hand, W being invariant under T is invariant
j
j
j
j
under every polynomial in T. In particular, W is invariant under the orthogonal projection E of
j
V on W . If is in W it follows that E is in W W , and, at the same time, = E .
j j j j j
j
Therefore, W is contained in W W .
j
j
Theorem 9 shows that every normal operator T on a finite-dimensional inner product space is
canonically specified by a finite number of normal operators T , defined on the primary
j
components W of V under T, each of whose minimal polynomials is irreducible over the field of
j
scalars. To complete our understanding of normal operators it is necessary to study normal
operators of this special type.
A normal operator whose minimal polynomial is of degree 1 is clearly just a scalar multiple of
the identity. On the other hand, when the minimal polynomial is irreducible and of degree 2 the
situation is more complicated.
Example 1: Suppose r > 0 and that is a real number which is not an integral multiple
2
of . Let T be the linear operator on R whose matrix in the standard orthonormal basis is
cos sin
A = r
sin cos
Then T is a scalar multiple of an orthogonal transformation and hence normal. Let p be the
characteristic polynomial of T. Then
p = det (xI – A)
2
= (x – r cos ) + r sin
2
2
2
= x – 2r cos x + r .
Let a = r cos , b = r sin , and c = a + ib. Then b 0, c = re i
a b
A =
b a
and p = (x – c)(x – c ). Hence p is irreducible over R. Since p is divisible by the minimal polynomial
for T, it follows that p is the minimal polynomial.
This example suggests the following converse.
Theorem 10: Let T be a normal operator on a finite-dimensional real inner product space V and p
its minimal polynomial. Suppose
p = (x – a) + b 2
2
8
where a and b are real and b 0. Then there is an integer s > 0 such that p is the characteristic
polynomial for T, and there exist subspaces V , …, V of V such that
1 s
(i) V is orthogonal to V when i j;
j i
(ii) V = V … V ;
1 s
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