14.6 The Right Polar Factorization∗
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 .
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 A1∕2 such that A1∕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 D1∕2 the matrix which is
obtained by replacing each diagonal entry of D with its square root. Thus D1∕2D1∕2 = D. Then
Since D1∕2 is real,
so A1∕2 is Hermitian. ■
In fact this square root is unique. This is shown a little later after the main result of this
Next it is helpful to recall the Gram Schmidt algorithm and observe a certain property stated in the
Lemma 14.6.2 Suppose
is a linearly independent set of vectors such
is an orthonormal set of vectors. Then when the Gram Schmidt process is applied
to the vectors in the given order, it will not change any of the w1,
be the orthonormal set delivered by the Gram Schmidt process. Then
because by definition, u1 ≡ w1∕
Now suppose uj
for all j ≤ k ≤ r.
Then if k < r,
consider the definition of uk+1.
By induction, uj = wj and so this reduces to wk+1∕
This lemma immediately implies the following lemma.
Lemma 14.6.3 Let V be a subspace of dimension p and let
be an orthonormal set of
vectors in V . Then this orthonormal set of vectors may be extended to an orthonormal basis for
Proof: First extend the given linearly independent set
to a basis for
and then apply
the Gram Schmidt theorem to the resulting basis. Since
is orthonormal it follows from
the result is of the desired form, an orthonormal basis extending
Recall Lemma 13.2.5 which is about preserving distances. It is restated here in the case of an m × n
Lemma 14.6.4 Suppose R is an m × n matrix with m ≥ n and R preserves distances. Then
R∗R = I.
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
Therefore, by Lemma 14.6.1, there exists an n×n Hermitian matrix U having all nonnegative eigenvalues
Consider the subspace U
be an orthonormal basis for
Note that U
might not be all of
Using Lemma 14.6.3
, extend to an orthonormal basis for all of
Next observe that
is also an orthonormal set of vectors in
. This is because
Therefore, from Lemma 14.6.3
again, this orthonormal set of vectors can be extended to an orthonormal
basis for Fm,
Thus there are at least as many zk as there are yj because m ≥ n. Now for x ∈ Fn, since
is an orthonormal basis for Fn, there exist unique scalars,
and so it follows from Corollary 13.2.8 or Lemma 14.6.4 that R∗R = I. Then also there exist scalars bk such
and so from 14.12,
and this shows
= Fx. ■
Note that U2 is completely determined by F because F∗F = UR∗RU = U2. In fact, U is also
uniquely determined. This will be shown later in Theorem 14.7.1. First is an easy corollary of this
Corollary 14.6.6 Let F be m×n and suppose n ≥ m. Then there exists a Hermitian U and andR, such
Proof: Recall that L∗∗ = L and
. Now apply Theorem 14.6.5
. Thus, F∗
satisfy the conditions of that theorem. In particular R∗
preserves distances. Then F