1.5. THE RIGHT POLAR FACTORIZATION 23

Theorem 1.5.5 Let F be an m×n matrix where m≥ n. Then there exists a Hermi-tian n× n matrix U which has all nonnegative eigenvalues and an m× n matrix R whichsatisfies R∗R = I such that F = RU.

Proof: Consider F∗F. This is a Hermitian matrix because

(F∗F)∗ = F∗ (F∗)∗ = F∗F

Also the eigenvalues of the n×n matrix F∗F are all nonnegative. This is because if x is aneigenvalue,

λ (x,x) = (F∗Fx,x) = (Fx,Fx)≥ 0.

Therefore, by Lemma 1.5.1, there exists an n×n Hermitian matrix U having all nonnegativeeigenvalues such that

U2 = F∗F.

Consider the subspace U (Fn). Let {Ux1, · · · ,Uxr} be an orthonormal basis for

U (Fn)⊆ Fn.

Note that U (Fn) might not be all of Fn. Using Lemma 1.5.3, extend to an orthonormalbasis for all of Fn,

{Ux1, · · · ,Uxr,yr+1, · · · ,yn} .

Next observe that {Fx1, · · · ,Fxr} is also an orthonormal set of vectors in Fm. This isbecause

(Fxk,Fx j) = (F∗Fxk,x j) =(U2xk,x j

)= (Uxk,U∗x j) = (Uxk,Ux j) = δ jk

Therefore, from Lemma 1.5.3 again, this orthonormal set of vectors can be extended to anorthonormal basis for Fm,

{Fx1, · · · ,Fxr,zr+1, · · · ,zm}

Thus there are at least as many zk as there are y j. Now for x ∈ Fn, since

{Ux1, · · · ,Uxr,yr+1, · · · ,yn}

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

c1 · · · ,cr,dr+1, · · · ,dn

such that

x=r

∑k=1

ckUxk +n

∑k=r+1

dkyk

Define

Rx≡r

∑k=1

ckFxk +n

∑k=r+1

dkzk (1.7)

Then also there exist scalars bk such that

U x=r

∑k=1

bkUxk