Page 313 - DMTH502_LINEAR_ALGEBRA

Unit 29: Spectral Theory and Properties of Normal Operators

r k Notes
n
= n c E j
j
n 0 j 1
k r
c
= n j n E j
j 1 n 0
r
c
= f ( )E j
j
j 1
Since e (c ) = , it follows that e (T) = E .
j m jm j j
Because E ,. ., E are canonically associated with T and
1 k
I = E + ... + E
1 k
the family of projections (E , ..., E ) is called the resolution of the identity defined by T.
1 k
There is a comment that should be made about the proof of the spectral theorem. We derived the
theorem using Theorems 9 and 13 of unit 26 on the diagonalization of self-adjoint and normal
operators. There is another, more algebraic, proof in which it must first be shown that the
minimal polynomial of a normal operator is a product of distinct prime factors. Then one
proceeds as in the proof of the primary decomposition theorem (Theorem 1) unit 18.
In various applications it is necessary to know whether one may compute certain functions of
operators or matrices, e.g., square roots. This may be done rather simply for diagonalizable
normal operators.
Definition: Let T be a diagonalizable normal operator on a finite-dimensional inner product
space and
k
T = c E
j j
j 1
its spectral resolution. Suppose f is a function whose domain includes the spectrum of T that has
values in the field of scalars. Then the linear operator f(T) is defined by the equation

k
E
c
f(T) = f ( ) . ...(2)
j
j
j 1
Theorem 2: Let T be a diagonalizable normal operator with spectrum S on a finite-dimensional
inner product space V. Suppose f is a function whose domain contains S that has values in the
field of scalars. Then f(T) is a diagonalizable normal operator with spectrum f(S). If U is a unitary
–1
map of V onto V' and T' = UTU , then S is the spectrum of T' and
f(T) = Uf(T)U .
–1
Proof: The normality of f(T) follows by a simple computation from (2) and the fact that
c
f(T)* = f ( )E j
j
j
Moreover, it is clear that for every in E (V)
j
f(T) = f(c ) .
j
Thus, the set f(S) of all f(c) with c in S is contained in the spectrum of f(T). Conversely, suppose
0 and that
f(T) = b .

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