Page 268 - Advanced Linear Algebra

252 Advanced Linear Algebra

Proof. Since is a positive operator, it has a positive square root j , which is

a polynomial in . A similar statement holds for . Therefore, since and
commute, so do j and j . Hence,
² ³ jj ~ ² ³ ² j ³ ~ j
Since j and j are self-adjoint and commute, their product is self-adjoint
and so is positive.
The Polar Decomposition of an Operator

It is well known that any nonzero complex number can be written in the polar
'

form '~ , where is a positive number and is real. We can do the same

for any nonzero linear operator on a finite-dimensional complex inner product

space.
Theorem 10.25 Let be a nonzero linear operator on a finite-dimensional

complex inner product space .
=
)
1 There exist a positive operator and a unitary operator for which

~ . Moreover, is unique and if is invertible, then is also unique.

)
2 Similarly, there exist a positive operator and a unitary operator for

which ~ . Moreover, is unique and if is invertible, then is also

unique.

Proof. Let us suppose for a moment that ~ . Then
i ~² i ³ ~ i i ~ c
and so
~ i c ~
Also, if #= , then
#~ ² #³

These equations give us a clue as to how to define and .

Let us define to be the unique positive square root of the positive operator

. Then
i
i
) ) # ~ º #Á #» ~ º #Á #» ~ º )~ # ) #Á #» (10.2 )

Define on im²³ by

²#³ ~ #
(
)

for all #= . Equation 10.2 shows that %~ & implies that % ~ & and so

this definition of on im²³ is well-defined.

)
(

Moreover, is an isometry on im²³ , since 10.2 gives