Polar Decomposition, Singular-Value Decomposition, and Autonne-Takagi Factorization

Polar decomposition, singular-value decomposition, and Autonne-Takagi factorization

Carlton M. Caves1, ∗ 1Center for Quantum Information and Control, University of New Mexico, Albuquerque NM 87131-0001, USA (Dated: February 14, 2017) This document provides an introduction to the polar decomposition, the singular-value decomposition, and Autonne-Takagi factorization of symmetric matrices, with an emphasis on how all of these are close to the same thing once one thinks about operators instead of their representations.

We begin by reviewing the polar decomposition and the singular-value decomposition, since Autonne-Takagi factorization is a special case of these. p Polar decomposition. Given a linear operator M, the polar decomposition starts with the positive operators M †M p † and MM . These two positive operators have the same (real, nonnegative) eigenvalues, denoted as λj. The † † 2 p † eigenvalues of M M and MM are the squares, λj . If we let |eji and |fji be (orthonormal) eigenvectors of M M p † and MM , respectively, with |eji and |fji having the same eigenvalue λj, we have

p † X M M = λj|ejihej| , (1) j p † X MM = λj|fjihfj| . (2) j

The eigenvalues λj are called the singular values of M, and the eigenvectors |eji and |fji are called the right and left singular vectors. Letting U be the unitary operator that transforms between the two eigenbases, i.e., X U|eji = |fji ⇐⇒ U = |fjihej| , (5) j we have that U transforms between the two positive operators, i.e., p p U M †MU † = MM † . (6)

This leads us to the polar decomposition, p p M = U M †M = MM †U , (7) p p M † = M †MU † = U † MM † . (8)

∗Electronic address: [email protected] 2

† 2 The unitary operator U is unique if M M is invertible (i.e., all the eigenvalues λj are nonzero). This is clear from the construction of the polar decomposition; moreover, when M †M is invertible, we have U = M(M †M)−1/2. p † p 2 For a Hermitian operator H, with eigenvalues hj and eigenvectors |eji, we have H H = H = |H|, i.e., λj = |hj|; the only role of the unitary U in the polar decomposition is to insert a sign change for negative eigenvalues of H: X U = sign (hj)|ejihej| . (9) j

A unitary operator U is its own polar decomposition. A normal operator N, i.e., N †N = NN †, has an eigendecom- P iφj position N = j λje |ejihej|, where the λjs are nonnegative, so

p † p † X N N = NN = λj|ejihej| ; (10) j the role of the unitary U in the polar decomposition is to put the phases back in:

X iφj U = e |ejihej| . (11) j p What distinguishes normal operators is that the unitary U and the positive operator N †N in the polar decomposition commute.

Singular-value decomposition. The singular-value decomposition is really the same thing as the polar decomposition, the only diﬀerence being that there is a standard basis |ji in which we want the singular values to appear in diagonal form as X Λ = λj|jihj| . (12) j

By introducing a unitary operator V that maps from the standard basis to the eigenbasis |eji, i.e., V |ji = |eji, we can write

p † X † M M = λj|ejihej| = V ΛV (13) j and thus put the polar decomposition in the form

M = W ΛV † , (14) where W = UV , with W |ji = |fji. Equation (14) is the singular-value decomposition. Notice that in this construction, we have the freedom to rephase the right singular vectors |eji or, equivalently, to multiply V on the right by a unitary that is diagonal in the standard basis. In the standard basis, the singular-value decomposition becomes

Autonne-Takagi factorization. Autonne-Takagi factorization is usually stated in terms of a complex symmetric matrix; if we wish to state it in terms of operators, we need a standard basis |ji relative to which transposition and complex conjugation are deﬁned. Now let’s state precisely what we want to prove:

If M = M T is a symmetric operator, relative to a standard basis |ji, there exists a unitary operator V such that

T X V MV = Λ = λj|jihj| , (16) j 3 where the diagonal elements are the (real, nonnegative) singular values of M.

Since M = V ∗ΛV †, this is a special case of the singular-value decomposition with W = V ∗. Translated to the polar decomposition, the unitary operator in the polar decomposition becomes U = WV † = V ∗V † = (VV T )†. Proof. Let S be a unitary operator such that S†M †MS = Λ2. Consider the symmetric operator L = ST MS = LT . One has L†L = S†M †S∗ST MS = S†M †MS = Λ2 and LL† = ST MSS†M †S∗ = ST MM †S∗ = (S†M †MS)∗ = Λ2. This implies that L is diagonal in the standard basis: X iφj X iφj X iφj /2 X iφj /2 L = λje |jihj| = e |jihj| Λ = e |jihj| Λ e |jihj| . (17) j j j j

The second form is the polar decomposition of the normal operator L. The last form is the one we now use by writing X −iφ /2 T X −iφ /2 Λ = e j |jihj| S MS e j |jihj| . (18) j j

Deﬁning X −iφ /2 V = S e j |jihj| (19) j completes the proof. Notice that to deﬁne V , what we used was the above-mentioned freedom to rephase the right singular vectors, so that V also diagonalizes MM †. Indeed, using

∗ † p † X ∗ M = V ΛV = U M M = λj|ej ihej| . (21) j

Thus one may think of the real content of Autonne-Takagi factorization as the fact that the left and right singular ∗ vectors of a symmetric operator can be rephased so that they are complex conjugates of one another, i.e., |fji = |ej i.