94 CHAPTER 5. SOME IMPORTANT LINEAR ALGEBRA

Then

0 = Re(R∗Rx−x,αy) = Reα (R∗Rx−x,y)= |(R∗Rx−x,y)|

Thus |(R∗Rx−x,y)|= 0 for all x,y because the given x,y were arbitrary. Let y = R∗Rx−xto conclude that for all x,

R∗Rx−x = 0

which says R∗R = I since x is arbitrary. This proves the lemma.With this preparation, here is the big theorem about the right polar decomposition.

Theorem 5.9.6 Let F be an m×n matrix where m≥ n. Then there exists a Hermitian n×nmatrix, U which has all nonnegative eigenvalues and an m× n matrix, R which preservesdistances and satisfies 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 5.9.1, there exists an n×n Hermitian matrix, U having all nonnega-tive eigenvalues 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 5.9.4, 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 5.9.4 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}