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The Spectral Theorem
                      for the details, the most important of which is verifying that T| {u} ⊥ is
                      self-adjoint (this allows us to apply our induction hypothesis).
                         Let λ be any eigenvalue of T (because T is self-adjoint, we know                  137
                      from the previous lemma that it has an eigenvalue) and let u ∈ V
                      denote a corresponding eigenvector with  u = 1. Let U denote the    To get an eigenvector
                      one-dimensional subspace of V consisting of all scalar multiples of u.  of norm 1, take any
                      Note that a vector v ∈ V is in U  ⊥  if and only if  u, v = 0.      nonzero eigenvector
                         Suppose v ∈ U . Then because T is self-adjoint, we have          and divide it by its
                                       ⊥
                                                                                          norm.
                                    u, Tv = Tu, v = λu, v = λ u, v = 0,
                      and hence Tv ∈ U . Thus Tv ∈ U  ⊥  whenever v ∈ U . In other words,
                                       ⊥
                                                                       ⊥
                      U ⊥  is invariant under T. Thus we can define an operator S ∈L(U ) by
                                                                                  ⊥
                      S = T| U ⊥.If v, w ∈ U , then
                                          ⊥
                                     Sv, w = Tv, w = v, Tw = v, Sw ,
                      which shows that S is self-adjoint (note that in the middle equality
                      above we used the self-adjointness of T). Thus, by our induction hy-
                      pothesis, there is an orthonormal basis of U  ⊥  consisting of eigenvec-
                      tors of S. Clearly every eigenvector of S is an eigenvector of T (because
                      Sv = Tv for every v ∈ U ). Thus adjoining u to an orthonormal basis
                                              ⊥
                      of U  ⊥  consisting of eigenvectors of S gives an orthonormal basis of V
                      consisting of eigenvectors of T, as desired.

                         For T ∈L(V) self-adjoint (or, more generally, T ∈L(V) normal
                      when F = C), the corollary below provides the nicest possible decom-
                      position of V into subspaces invariant under T. On each null(T − λ j I),
                      the operator T is just multiplication by λ j .

                      7.14   Corollary: Suppose that T ∈L(V) is self-adjoint (or that F = C
                      and that T ∈L(V) is normal). Let λ 1 ,...,λ m denote the distinct eigen-
                      values of T. Then
                                    V = null(T − λ 1 I) ⊕· · ·  null(T − λ m I).
                      Furthermore, each vector in each null(T − λ j I) is orthogonal to all vec-
                      tors in the other subspaces of this decomposition.

                         Proof:   The spectral theorem (7.9 and 7.13) implies that V has a
                      basis consisting of eigenvectors of T. The desired decomposition of V
                      now follows from 5.21.
                         The orthogonality statement follows from 7.8.
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