5.9. THE RIGHT POLAR DECOMPOSITION 95

is an orthonormal basis for Fn, there exist unique scalars,

c1 · · · ,cr,dr+1, · · · ,dn

such that

x =r

∑k=1

ckUxk +n

∑j=r+1

dkyk

Define

Rx≡r

∑k=1

ckFxk +n

∑j=r+1

dkzk (5.9.27)

Then also there exist scalars bk such that

Ux =r

∑k=1

bkUxk

and so from 5.9.27,applied to Ux in place of x

RUx =r

∑k=1

bkFxk = F

(r

∑k=1

bkxk

)

Is F (∑rk=1 bkxk) = F (x)?(

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.

From 5.9.27 and Lemma 5.9.2 R preserves distances. Therefore, by Lemma 5.9.5 R∗R = I.This proves the theorem.

Here is a useful fact from Linear algebra.