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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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