13.4. THE RIGHT POLAR FACTORIZATION∗ 315

By induction, u j = w j and so this reduces to wk+1/ |wk+1|= wk+1. ■This lemma immediately implies the following lemma.

Lemma 13.4.3 Let V be a subspace of dimension p and let {w1, · · · ,wr} be an orthonor-mal set of vectors in V . Then this orthonormal set of vectors may be extended to an or-thonormal basis for V, {

w1, · · · ,wr,yr+1, · · · ,yp}

Proof: First extend the given linearly independent set {w1, · · · ,wr} to a basis for Vand then apply the Gram Schmidt theorem to the resulting basis. Since {w1, · · · ,wr} isorthonormal it follows from Lemma 13.4.2 the result is of the desired form, an orthonormalbasis extending {w1, · · · ,wr}. ■

Here is another lemma about preserving distance.

Lemma 13.4.4 Suppose R is an m×n matrix with m≥ n and R preserves distances. ThenR∗R = I.

Proof: Since R preserves distances, |Rx| = |x| for every x. Therefore from the axiomsof the dot product,

|x|2 + |y|2 +(x,y)+(y,x) = |x+y|2 = (R(x+y) ,R(x+y))= (Rx,Rx)+(Ry,Ry)+(Rx,Ry)+(Ry,Rx)= |x|2 + |y|2 +(R∗Rx,y)+(y,R∗Rx)

and so for all x,y,(R∗Rx−x,y)+(y,R∗Rx−x) = 0

Hence for all x,y,Re(R∗Rx−x,y) = 0

Now for a x,y given, choose α ∈ C such that

α (R∗Rx−x,y) = |(R∗Rx−x,y)|

Then0 = 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. ■With this preparation, here is the big theorem about the right polar factorization.

Theorem 13.4.5 Let F be an m×n matrix where m≥ n. Then there exists a Hermitian n×n matrix U which has all nonnegative eigenvalues and an m×n matrix R which preservesdistances and satisfies R∗R = I such that

F = RU.