Page 165 - Linear Algebra Done Right

Polar and Singular-Value Decompositions
where the first factor, namely, z/|z|, is an element of the unit circle. Our
analogy leads us to guess that any operator T ∈L(V) can be written
as an isometry times √ T T. That guess is indeed correct, as we now 153
∗
prove.
7.41 Polar Decomposition: If T ∈L(V), then there exists an isom- If you know a bit of
etry S ∈L(V) such that complex analysis, you
√
T = S T T. will recognize the
∗
analogy to polar
Proof: Suppose T ∈L(V).If v ∈ V, then coordinates for
complex numbers:
2
Tv = Tv, Tv
every complex number
∗
= T Tv, v can be written in the
√ √ θi
∗
∗
= T T T Tv, v form e r, where
√ √ θ ∈ [0, 2π) and r ≥ 0.
∗
∗
= T Tv, T Tv θi
√ Note that e is in the
2
= T Tv . unit circle,
∗
corresponding to S
Thus
being an isometry, and
√
7.42 Tv = T Tv r is nonnegative,
∗
corresponding to
√
for all v ∈ V. T T being a positive
∗
√
Define a linear map S 1 : range T T → range T by operator.
∗
√
7.43 S 1 ( T Tv) = Tv.
∗
The idea of the proof is to extend S 1 to an isometry S ∈L(V) such that
√
T = S T T. Now for the details.
∗
First we must check that S 1 is well defined. To do this, suppose
√ √
v 1 ,v 2 ∈ V are such that T Tv 1 = T Tv 2 . For the definition given
∗
∗
by 7.43 to make sense, we must show that Tv 1 = Tv 2 . However,
Tv 1 − Tv 2 = T(v 1 − v 2 )
√
∗
= T T(v 1 − v 2 )
√ √
∗
∗
= T Tv 1 − T Tv 2
= 0,
where the second equality holds by 7.42. The equation above shows
that Tv 1 = Tv 2 ,so S 1 is indeed well defined. You should verify that S 1
is a linear map.