Page 283 - DMTH502_LINEAR_ALGEBRA

Unit 26: Unitary Operators and Normal Operators

V onto W not only to preserve the linear operations, but also to preserve products. An Notes
isomorphism of an inner product space onto itself is called a ‘unitary operator’ on that space. Some
of the basic properties of unitary operators are being established in the section along with some
examples.
Definition: Let V and W be inner product spaces over the same field and let T be a linear
transformation from V onto W. We say that T-preserves inner products if (T \T ) = ( \ ) for all
, in V. An isomorphism of V onto W is a vector space isomorphism T of V onto W which also
preserves inner products.
If T preserves inner products then T = and so T is non-singular. Thus if T is an isomorphism
–1
of V onto W, then T is an isomorphism of W onto V; hence, when such a T exists, we shall simply
say V and W are isomorphic. Of course, isomorphism of inner product spaces is an equivalence
relation.
Theorem 1: Let V and W be finite-dimensional inner product spaces over the same field, having
the same dimension. If T is a linear transformation from V into W, the following are equivalent.

(i) T preserves inner products.
(ii) T is an (inner product space) isomorphism.
(iii) T carries every orthonormal basis for V onto an orthonormal basis for W.
(iv) T carries some orthonormal basis for V onto an orthonormal basis for W.

Proof: (i) (ii) If T preserves inner products, then T = for all in V. Thus T is non-
singular, and since dim V = dim W, we know that T is a vector space isomorphism.
(ii) (iii) Suppose T is an isomorphism. Let { , …, } be an orthonormal basis for V. Since T is
1 n
a vector space isomorphism and dim W = dim V, it follows that {T , …, T } is a basis for W.
1 n
Since T also preserves inner products, {T |T } = ( | ) = .
1 k j k jk
(iii) (iv) This requires no comment.
(iv) (i) Let { , …, } be an orthonormal basis for V such that {T , …, T } is an orthonormal
1 n 1 n
basis for W. Then
(T |T ) = ( | ) = .
j k j k jk
For any = x + … + x and = y + … + y in V, we have
1 1 n n 1 1 n n
n
( | ) = x y j
j
j 1

(T |T ) = x T j y T k
k
j
j k
= x y k (T j |T k )
j
j k
n
= x y j
j
j 1
and so T preserves inner products.

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