13.7. THE RIGHT POLAR FACTORIZATION 353

such that

x=r

∑k=1

ckUxk +n

∑k=r+1

dkyk

Define

Rx≡r

∑k=1

ckFxk +n

∑k=r+1

dkzk (13.13)

Thus, since {Fx1, · · · ,Fxr,zr+1, · · · ,zn} is orthonormal,

|Rx|2 =r

∑k=1|ck|2 +

n

∑k=r+1

|dk|2 = |x|2

and so it follows from Corollary 12.3.8 or Lemma 13.7.4 that R∗R = I. Then also thereexist scalars bk such that

U x=r

∑k=1

bkUxk (13.14)

and so from 13.13,

RU x=r

∑k=1

bkFxk = F

(r

∑k=1

bkxk

)Is F (∑r

k=1 bkxk) = F (x)? Using 13.14,(F

(r

∑k=1

bkxk

)−F (x) ,F

(r

∑k=1

bkxk

)−F (x)

)

=

((F∗F)

(r

∑k=1

bkxk−x

),

(r

∑k=1

bkxk−x

))

=

(U2

(r

∑k=1

bkxk−x

),

(r

∑k=1

bkxk−x

))

=

(U

(r

∑k=1

bkxk−x

),U

(r

∑k=1

bkxk−x

))

=

(r

∑k=1

bkUxk−Ux,r

∑k=1

bkUxk−Ux

)= 0

Therefore, F (∑rk=1 bkxk) = F (x) and this shows RUx= Fx. ■

Note that U2 is completely determined by F because F∗F =UR∗RU =U2. In fact, Uis also uniquely determined. This will be shown later in Theorem 13.8.1. First is an easycorollary of this theorem.

Corollary 13.7.6 Let F be m×n and suppose n≥m. Then there exists a Hermitian U andand R, such that

F =UR, RR∗ = I.

Proof: Recall that L∗∗ = L and (ML)∗ = L∗M∗. Now apply Theorem 13.7.5 to F∗.Thus, F∗ = R∗U where R∗ and U satisfy the conditions of that theorem. In particular R∗

preserves distances. Then F =UR and RR∗ = R∗∗R∗ = I. ■