Page 285 - DMTH502_LINEAR_ALGEBRA

Unit 26: Unitary Operators and Normal Operators

= (PY)*(PX) Notes
= Y*P*PX
= Y*GX
= [X|Y].

Hence T is an isomorphism.

Example 4: Let V be the space of all continuous real-valued functions on the unit interval,
0 t 1, with the inner product

1
2
[f|g] = f ( ) ( )t dt .
g
t
t
0
Let W be the same vector space with the inner product
1
(f|g) = f ( ) ( ) dt .
t
g
t
0
Let T be the linear transformation from V into W given by
(Tf) (t) = tf(t).
Then (Tf|Tg) = [f|g], and so T preserves inner products; however, T is not an isomorphism of V
onto W, because the range of T is not all of W. Of course, this happens because the underlying
vector space is not finite dimensional.
Theorem 2: Let V and W be inner product spaces over the same field, and let T be a linear
transformation from V into W. Then T preserves inner products if and only if T = for
every in V.
Proof: If T preserves inner products, T ‘preserves norms’. Suppose T = for every in V.
2
2
Then T = . Now using the appropriate polarization identity and the fact that T is linear,
one easily obtains ( | ) = (T |T ) for all , in V.
Definition: A unitary operator on an inner product space is an isomorphism of the space onto
itself.
The product of two unitary operators is unitary. For, if U and U are unitary, then U U is
1 2 2 1
invertible and U U = U = for each . Also, the inverse of a unitary operator is
2 1 1
unitary, since U = says U –1 = , where = U . Since the identity operator is clearly
unitary, we see that the set of all unitary operators on an inner product space is a group, under
the operation of composition.

If V is a finite-dimensional inner product space and U is a linear operator on V, Theorem 1 tells
us that U is unitary if and only if (U |U ) = ( | ) for each , in V; or, if and only if for some
(every) orthonormal basis { , …, } it is true that {U , …, U } is an orthonormal basis.
1 n 1 n
Theorem 3: Let U be a linear operator on an inner product space V. Then U is unitary if and only
if the adjoint U* of U exists and UU* = U*U = I.

Proof: Suppose U is unitary. Then U is invertible and
(U | ) = (U |UU –1 ) = ( |U –1 )
for all , . Hence U is the adjoint of U.
–1
Conversely, suppose U* exists and UU* = U*U = I. Then U is invertible, with U = U*. So, we need
–1
only show that U preserves inner products.

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