The right polar factorization involves writing a matrix as a product of two other matrices, one which preserves distances and the other which stretches and distorts. This is of fundamental significance in geometric measure theory and also in continuum mechanics. Not surprisingly the stress should depend on the part which stretches and distorts. See [14].
First here are some lemmas which review and add to many of the topics discussed so far about adjoints and orthonormal sets and such things.
Lemma 14.6.1 Let A be a Hermitian matrix such that all its eigenvalues are nonnegative. Then there exists a Hermitian matrix A^{1∕2} such that A^{1∕2} has all nonnegative eigenvalues and
Proof: Since A is Hermitian, there exists a diagonal matrix D having all real nonnegative entries and a unitary matrix U such that A = U^{∗}DU. Then denote by D^{1∕2} the matrix which is obtained by replacing each diagonal entry of D with its square root. Thus D^{1∕2}D^{1∕2} = D. Then define

Then

Since D^{1∕2} is real,

so A^{1∕2} is Hermitian. ■
In fact this square root is unique. This is shown a little later after the main result of this section.
Next it is helpful to recall the Gram Schmidt algorithm and observe a certain property stated in the next lemma.
Proof: Let

By induction, u_{j} = w_{j} and so this reduces to w_{k+1}∕
This lemma immediately implies the following lemma.
Lemma 14.6.3 Let V be a subspace of dimension p and let

Proof: First extend the given linearly independent set
Recall Lemma 13.2.5 which is about preserving distances. It is restated here in the case of an m × n matrix.
With this preparation, here is the big theorem about the right polar factorization.
Theorem 14.6.5 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 preserves distances and satisfies R^{∗}R = I such that F = RU.
Proof: Consider F^{∗}F. This is a Hermitian matrix because

Also the eigenvalues of the n × n matrix F^{∗}F are all nonnegative. This is because if x is an eigenvalue,

Therefore, by Lemma 14.6.1, there exists an n×n Hermitian matrix U having all nonnegative eigenvalues such that

Consider the subspace U

Note that U

Next observe that

Thus there are at least as many z_{k} as there are y_{j} because m ≥ n. Now for x ∈ F^{n}, since

is an orthonormal basis for F^{n}, there exist unique scalars,

such that

Define
 (14.12) 
Thus, since

and so it follows from Corollary 13.2.8 or Lemma 14.6.4 that R^{∗}R = I. Then also there exist scalars b_{k} such that
 (14.13) 
and so from 14.12,

Is F

Note that U^{2} is completely determined by F because F^{∗}F = UR^{∗}RU = U^{2}. In fact, U is also uniquely determined. This will be shown later in Theorem 14.7.1. First is an easy corollary of this theorem.
Corollary 14.6.6 Let F be m×n and suppose n ≥ m. Then there exists a Hermitian U and andR, such that

Proof: Recall that L^{∗∗} = L and