Linear Algebra, Theory And Applications, 2012a

13.6. POLAR DECOMPOSITIONS 323
Proof: (F ∗ F ) ∗ = F ∗ F and so by Theorem 13.3.3, there is an orthonormal basis of
eigenvectors, {v 1 , ··· , v n } such that
It is also clear that λ i ≥ 0 because
Let
F ∗ F v i = λ i v i ,F ∗ F =
n∑
λ i v i ⊗ v i .
i=1
λ i (v i , v i )=(F ∗ F v i , v i )=(F v i ,Fv i ) ≥ 0.
U ≡
n∑
i=1
λ 1/2
i v i ⊗ v i .
{
Then U 2 = F ∗ F, U = U ∗ , and the eigenvalues of U,
λ 1/2
i
} n
i=1
are all non negative.
Let {Ux 1 , ··· ,Ux r } be an orthonormal basis for U (X) . By the Gram Schmidt procedure
there exists an extension to an orthonormal basis for X,
{Ux 1 , ··· ,Ux r , y r+1 , ··· , y n } .
Next note that {F x 1 , ··· ,Fx r } is also an orthonormal set of vectors in Y because
(F x k ,Fx j )=(F ∗ F x k , x j )= ( U 2 x k , x j
)
=(Uxk ,Ux j )=δ jk .
By the Gram Schmidt procedure, there exists an extension of {F x 1 , ··· ,Fx r } to an orthonormal
basis for Y,
{F x 1 , ··· ,Fx r , z r+1 , ··· , z m } .
Since m ≥ n, there are at least as many z k as there are y k .Nowforx ∈ X, since
{Ux 1 , ··· ,Ux r , y r+1 , ··· , y n }
is an orthonormal basis for X, there exist unique scalars
such that
Define
Thus
x =
Rx ≡
|Rx| 2 =
c 1 , ··· ,c r ,d r+1 , ··· ,d n
r∑
c k Ux k +
k=1
r∑
c k F x k +
k=1
r∑
|c k | 2 +
k=1
Therefore, by Lemma 13.6.1 R ∗ R = I.
Then also there exist scalars b k such that
Ux =
n∑
k=r+1
n∑
k=r+1
n∑
k=r+1
d k y k
d k z k (13.18)
|d k | 2 = |x| 2 .
r∑
b k Ux k (13.19)
k=1