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Linear Algebra

Notes 29.1 Spectral Theory

In this unit we try to implement the findings of the Theorems 9 and 13 of unit 26 regarding the
diagonalization of self-adjoint and normal operators.

We start with the following spectral theorem:
Theorem 1 (Spectral Theorem): Let T be a normal operator on a finite dimensional complex inner
product space V or a self-adjoint operator on a finite dimensional real inner product space. Let
C , ... C be the distinct characteristic values of T. Let W be the characteristic space associated with
1 k j
C and E , the orthogonal projection of V on W . Then W is orthogonal to W* when i j, V is the
j j j i j
direct sum of W , W , ... W and
1 2 k
T = C E + C E + ... + C E ...(1)
1 1 2 2 k k
Proof: Let be a vector in W , a vector in W , and suppose i j. Then c,( | ) = (T | ) =
j i
( |T* ) = ( |c i ). Hence (c – c )( | ) = 0, and since. c – c 0, it follows that ( | ) = 0. Thus W j
j
i
i
j
is orthogonal to W , when i j. From the fact that V has an orthonormal basis consisting of
i
characteristic vectors (cf. Theorems 9 and 13 of unit 26), it follows that V = W + ... + W . If
1 k j
belongs to V (1 j k) and + ... + = 0, then
j 1 k
0 = ( | j ) ( | j )
i
i
j j
2
= i
for every i, so that V is the direct sum of W , ... , W . Therefore E + ... + E = I and
1 k 1 k
T = TE + ... + TE .
1 k
= c E + ... + c E
1 1 k k
The decomposition (1) is called the spectral resolution of T. This terminology arose in part from
physical applications which caused the spectrum of a linear operator on a finite-dimensional
vector space to be defined as the set of characteristic values for the operator. It is important to
note that the orthogonal projections E , ... , E are canonically associated with T; in fact, they are
1 k
polynomials in T.
x c i
Corollary: If e = II , then E = e (T) for 1 j k.
j j j
i j c j c i
Proof: Since E E = 0 when i j, it follows that
i j
2
2
T = c E + ... + c E
2
1 1 k k
and by all easy induction argument that
T = c E + ... + c E
n
n
n
1 1 k k
for every integer n 0. For an arbitrary polynomial
r n
f = n x
n 0
we have
r n
f(T) = n T
n 0

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