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Unit 26: Unitary Operators and Normal Operators
For, any two diagonal matrices commute, and since T and T* are both represented by diagonal Notes
matrices in the ordered basis , we have (2). It is a rather remarkable fact that in the complex case
this condition is also sufficient to imply the existence of an orthonormal basis of characteristic
vectors.
Definition: Let V be a finite-dimensional inner product space and T a linear operator on V. We
say that T is normal if it commutes with its adjoint i.e., TT* = T*T.
Any self-adjoint operator is normal, as is any unitary operator. Any scalar multiple of a normal,
operator is normal; however, sums and products of normal operators are not generally normal.
Although it is by no means necessary, we shall begin our study of normal operators by considering
self-adjoint operators.
Theorem 6: Let V be an inner product space and T a self-adjoint linear operator on V. Then each
characteristic value of T is real, and characteristic vectors of T associated with distinct characteristic
values are orthogonal.
Proof: Suppose c is a characteristic value of T, i.e., that T = c for some non-zero vector . Then
c(|) = (c|)
= (T|)
= (|T)
= (|c)
= ( | )c
Since (|) 0, we must have c . c Suppose we also have T = d with 0. Then
c(|) = (T|)
= (|T)
= (|d)
= ( | )d
= d(|)
If c d, then (|) = 0.
It should be pointed out that Theorem 6 says nothing about the existence of characteristic values
or characteristic vectors.
Theorem 7: On a finite-dimensional inner product space of positive dimension, every self-adjoint
operator has a (non-zero) characteristic vector.
Proof: Let V be an inner product space of dimension n, where n > 0, and let T be a self-adjoint
operator on V. Choose an orthonormal basis for V and let A = [T] . Since T = T*, we have
A = A*. Now let W be the space of n 1 matrices over C, with inner product (X|Y) = Y*X. Then
U(X) = AX defines a self-adjoint linear operator U on W. The characteristic polynomial, det
(xI – A), is a polynomial of degree n over the complex numbers; every polynomial over C of
positive degree has a root. Thus, there is a complex number c such that det (cI – A) = 0. This means
that A – cI is singular, or that there exists a non-zero X such that AX = cX. Since the operator U
(multiplication by A) is self-adjoint, it follows from Theorem 6 that c is real. If V is a real vector
space, we may choose X to have real entries. For then A and A – cI have real entries, and since
A – cI is singular, the system (A – cI)X = 0 has a non-zero real solution X. It follows that there is
a non-zero vector in V such that T = c.
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