Page 290 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes for every in V. If U is the unique linear operator on V defined by U = , then P is the matrix
j j
of U in the ordered basis B:
n
= P j .
k jk
j 1
Since and are orthonormal bases, U is a unitary operator and P is a unitary matrix. If T is any
linear operator on V, then
–1
[T] = P [T] P = P*[T] P.
Definition: Let A and B be complex n × n matrices. We say that B is unitarily equivalent to A if
there is an n × n unitary matrix P such that B = P AP. We say that B is orthogonally equivalent
–1
to A if there is an n × n orthogonal matrix P such that B = P AP.
–1
With this definition, what we observed above may be stated as follows: If and are two
ordered orthonormal bases for V, then, for each linear operator T on V, the matrix [T] is

unitarily equivalent to the matrix [T] . In case V is a real inner product space, these matrices are

orthogonally equivalent, via a real orthogonal matrix.
Self Assessment

1. Let B given by

3 0 4
B = 1 0 7
2 9 11

is 3 × 3 invertible matrix. Show that there exists a unique lower triangular matrix M with
positive entries on the main diagonal such that MB is unitary. Find M and MB.
2. Let V be a complex inner product space and T a self-adjoint linear operator on V. Show that
(i) I + i T is non-singular

(ii) I – i T is non-singular
–1
(iii) (I – i T) (I + i T) is unitary.
26.2 Normal Operators

In this section we are interested in finding out the fact that there is an orthonormal basis for V
such that the matrix of the linear operator T on a finite dimensional inner product space V, in the
basis is diagonal.
We shall begin by deriving some conditions on T which will be subsequently shown to be
sufficient. Suppose = ( , …, ) is an orthonormal basis for V with the property
1 n
T = C a , j = 1, 2, … n … (1)
j j j
This simply says that T in this ordered basis is a diagonal matrix with diagonal entries c , c , …
1 2
c . If V is a real inner product space, the scalars c , …, c are (of course) real, and so it must be that
n 1 n
T = T*. In other words, if V is a finite-dimensional real inner product space and T is a linear
operator for which there is an orthonormal basis of characteristic vectors, then T must be
self-adjoint. If V is a complex inner product space, the scalars c , …, c need not be real, i.e., T need
1 n
not be self-adjoint. But notice that T must satisfy
TT* = T*T. … (2)

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