13.7. THE RIGHT POLAR FACTORIZATION 351

13.7 The Right Polar FactorizationThe 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 fun-damental significance in geometric measure theory and also in continuum mechanics. Notsurprisingly the stress should depend on the part which stretches and distorts. See [18].

First here are some lemmas which review and add to many of the topics discussed sofar about adjoints and orthonormal sets and such things.

Lemma 13.7.1 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 eigenvaluesand

(A1/2

)2= A.

Proof: Since A is Hermitian, there exists a diagonal matrix D having all real non-negative entries and a unitary matrix U such that A = U∗DU. Then denote by D1/2 thematrix which is obtained by replacing each diagonal entry of D with its square root. ThusD1/2D1/2 = D. Then define

A1/2 ≡U∗D1/2U.

Then (A1/2

)2=U∗D1/2UU∗D1/2U =U∗DU = A.

Since D1/2 is real, (U∗D1/2U

)∗=U∗

(D1/2

)∗(U∗)∗ =U∗D1/2U

so A1/2 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.

Lemma 13.7.2 Suppose{w1, · · · ,wr,vr+1, · · · ,vp

}is a linearly independent set of vec-

tors such that {w1, · · · ,wr} is an orthonormal set of vectors. Then when the Gram Schmidtprocess is applied to the vectors in the given order, it will not change any of the w1, · · · ,wr.

Proof: Let{u1, · · · ,up

}be the orthonormal set delivered by the Gram Schmidt pro-

cess. Then u1 =w1 because by definition, u1 ≡w1/ |w1| =w1. Now suppose u j =w jfor all j ≤ k ≤ r. Then if k < r, consider the definition of uk+1.

uk+1 ≡wk+1−∑

k+1j=1 (wk+1,u j)u j∣∣∣wk+1−∑k+1j=1 (wk+1,u j)u j

∣∣∣By induction, u j =w j and so this reduces to wk+1/ |wk+1|=wk+1 since |wk+1|= 1. ■

This lemma immediately implies the following lemma.

Lemma 13.7.3 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}

13.7. THE RIGHT POLAR FACTORIZATION 35113.7. The Right Polar FactorizationThe 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 fun-damental significance in geometric measure theory and also in continuum mechanics. Notsurprisingly the stress should depend on the part which stretches and distorts. See [18].First here are some lemmas which review and add to many of the topics discussed sofar about adjoints and orthonormal sets and such things.Lemma 13.7.1 Let A be a Hermitian matrix such that all its eigenvalues are nonnegative.Then there exists a Hermitian matrix A'/? such that A‘/? has all nonnegative eigenvaluesand (al/2)* =A.Proof: Since A is Hermitian, there exists a diagonal matrix D having all real non-negative entries and a unitary matrix U such that A = U*DU. Then denote by D'/? thematrix which is obtained by replacing each diagonal entry of D with its square root. ThusD!/2p!/2 — D. Then defineAl? =u*D!/7U.Then 5(4"?) =U*D'/2UU*D'/?U = U*DU =A.Since D!/? is real,(uD'7u) =u" (v7) (u*)* =u*D'2Uso A!/? is Hermitian.In fact this square root is unique. This is shown a little later after the main result of thissection.Next it is helpful to recall the Gram Schmidt algorithm and observe a certain propertystated in the next lemma.Lemma 13.7.2 Suppose {wy yt Wry Urtis ,Up} is a linearly independent set of vec-tors such that {w,,--- ,w,} is an orthonormal set of vectors. Then when the Gram Schmidtprocess is applied to the vectors in the given order, it will not change any of the w1,--+ , Wr.Proof: Let {u yo ,up} be the orthonormal set delivered by the Gram Schmidt pro-cess. Then wu; = wy, because by definition, w; = w)/|w | = w1. Now suppose uj = w;for all j <k <r. Then if k < r, consider the definition of wz+1.k+lWk+l — ea (wey1,Uj) UjUk+1 =k+1wes —Vi (Wr41,U;) UjBy induction, w; = w; and so this reduces to wy41/|wey1| = We41 Since |wy41|=1. HtThis lemma immediately implies the following lemma.Lemma 13.7.3 Let V be a subspace of dimension p and let {wy ,--- ,w,} 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,{w+ »WryUrtis Vp}