Page 315 - DMTH502_LINEAR_ALGEBRA

Unit 29: Spectral Theory and Properties of Normal Operators

In thinking about the preceding discussion, it is important for one to keep in mind that the Notes
spectrum of the normal operator T is the set
S ={c , ... , c }
1 k
of distinct characteristic values. When T is represented by a diagonal matrix in a basis of
characteristic vectors, it is necessary to repeat each value c as many times as the dimension of the
j
corresponding space of characteristic vectors. This is the reason for the change of notation in the
following result.
Corollary: With the assumptions of Theorem 2, suppose that T is represented in the ordered
basis = { , ... , } by the diagonal matrix D with entries d , ... , d . Then, in the basis , f(T) is
1 n 1 n
represented by the diagonal matrix f(D) with entries f(d ), ... , f(d ). If ’ = { , ... , } is any other
1 n 1 n
ordered basis and P the matrix such that
P
= ij i
j
j
–1
then P f(D)P is the matrix of f(T) in the basis '.
Proof: For each index i, there is a unique j such that 1 j k, belongs to E (V), and d = c . Hence
i j i j
f(T) = f(d ) for every i, and
i i i
P f ( )
T
f(T) = ij i
j
j
d P
= i ij i
j
(DP )
= ij i
j
(DP ) P 1
= ij ki k
j k
1
(P DP ) .
= kj k
k
It follows from this result that one may form certain functions of a normal matrix. For suppose
–1
A is a normal matrix. Then there is an invertible matrix P, in fact a unitary P, such that PAP is
a diagonal matrix, say D with entries d , ..., d ; Let f be a complex-valued function which can be
1 n
applied to d , ... d , and let f(D) be the diagonal matrix with entries f(d ) .....f(d ). Then P f(D)P is
–1
1 n 1 n
independent of D and just a function of A in the following sense. If Q is another invertible matrix
–1
such that QAQ is a diagonal matrix D’, then f may be applied to the diagonal entries of D’ and
P f(D)P = Q f(D’)Q.
–1
–1
–1
Definition: Under the above conditions, f(A) is defined as P f(D)P.
Theorem 3: Let A be a normal matrix and c , ..., c , the distinct complex roots of det (xl – A). Let
1 k
x c j
e = II
i j i c i c j
and E = e (A) (1 i k). Then E E = 0 when i j, E = E , E* = E ,
2
i i i j 1 i i i
and
I = E + ... + E .
1 k

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