Polar decomposition

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In mathematics, particularly in linear algebra and functional analysis, the polar decomposition of a matrix or linear operator is a factorization analogous to the polar form of a nonzero complex number z

z = r e^{i \theta}\,

where r is the absolute value of z (a positive real number), and eiθ is called the complex sign of z.

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[edit] Matrix polar decomposition

The polar decomposition of a complex matrix A is a matrix decomposition of the form

A = UP\,

where U is a unitary matrix and P is a positive-semidefinite Hermitian matrix. This decomposition always exists; and so long as A is invertible, it is unique, with P positive-definite. Note that

\det A = \det P\,\det U = re^{i\theta}

gives the corresponding polar decomposition of the determinant of A, since detP = r = | detA | and detU = eiθ.
The matrix P is given by

P = \sqrt{A^*A}

where A* denotes the conjugate transpose of A. This expression is meaningful since a positive-definite Hermitian matrix has a unique positive square root. The matrix U is then given by

U = AP^{-1}.\,

In terms of the singular value decomposition of A, A = W Σ V*, one has

P = V \Sigma V^*\,
U = W V^*\,

confirming that P is positive-definite and U is unitary.

One can also decompose A in the form

A = P'U\,

Here U is the same as before and P′ is given by

P' = UPU^{-1} = \sqrt{AA^*} = W \Sigma W^*.

This is known as the left polar decomposition, whereas the previous decomposition is known as the right polar decomposition. Left polar decomposition is also known as reverse polar decomposition.

The matrix A is normal if and only if P′ = P. Then UΣ = ΣU, and it is possible to diagonalise U with a unitary similarity matrix S that commutes with Σ, giving S U S* = Φ-1, where Φ is a diagonal unitary matrix of phases eiφ. Putting Q = V S*, one can then re-write the polar decomposition as

A = (Q \Phi Q^*)(Q \Sigma Q^*),\,

so A then thus also has a spectral decomposition

A = Q \Lambda Q^* \,

with complex eigenvalues such that ΛΛ* = Σ2 and a unitary matrix of complex eigenvectors Q.

The map from the general linear group GL(n,C) to the unitary group U(n) defined by mapping A onto its unitary piece U gives rise to a homotopy equivalence since the space of positive-definite matrices is contractible. In fact U(n) is the maximal compact subgroup of GL(n,C).

[edit] Bounded operators on Hilbert space

The polar decomposition of any bounded linear operator A between complex Hilbert spaces is a canonical factorization as the product of a partial isometry and a non-negative operator.

The polar decomposition for matrices generalizes as follows: if A is a bounded linear operator then there is a unique factorization of A as a product A = UP where U is a partial isometry, P is a non-negative self-adjoint operator and the initial space of U is the closure of the range of P.

We can see why U must be weakened to a partial isometry, rather than unitary, by considering a simple example. Let A be the one-sided shift on l2(N). So |A| = {A*A}½ = I. If U is such that A = U |A|, U must be A, which is not unitary.

We now prove the existence of polar decomposition. It is a consequence of the following simple fact:

Lemma If A, B are bounded operators on a Hilbert space H, and A*A ≤ B*B, then there exists a contraction C such that A = CB. Furthermore, C is unique if Ker(B*) ⊂ Ker(C).

proof: Define an operator C by C(Bh) = Ah, then extend by continuity to closure of Ran(B) and by zero to all of H. Because A*A ≤ B*B implies Ker(A) ⊂ Ker(B), C has the desired properties. This proves the lemma.

In particular. If A*A = B*B, then C is a partial isometry, which is unique if Ker(B*) ⊂ Ker(C). In general, for any bounded operator A,

A^*A = (A^*A)^{\frac{1}{2}} (A^*A)^{\frac{1}{2}},

where (A*A)½ is the unique positive square root of A*A given by the Borel functional calculus. So by lemma, we have

A = U (A^*A)^{\frac{1}{2}}

for some partial isometry U, which is unique if Ker(A*) ⊂ Ker(U). Take P to be (A*A)½ and one obtains the polar decomposition A = UP. Notice that an analogous argument can be used to show A = P'U' , where P' is positive and U' a partial isometry.

When H is finite dimensional, U can be extended to an unitary operator; this is not true in general (see example above). Alternatively, the polar decomposition can be shown using the operator version of singular value decomposition.

By property of the continuous functional calculus, |A| is in the C*-algebra generated by A. A similar but weaker statement holds for the partial isometry: U is in the von Neumann algebra generated by A. If A is invertible, the polar part U will be in the C*-algebra as well.

[edit] Unbounded operators

If A is a closed, densely defined unbounded operator between complex Hilbert spaces then it still has a (unique) polar decomposition

A = U |A|\,

where |A| is a (possibly unbounded) non-negative self adjoint operator with the same domain as A, and U is a partial isometry vanishing on the orthogonal complement of the range Ran(|A|).

The proof uses the same lemma as above, which goes through for unbounded operators in general. If Dom(A*A) = Dom(B*B) and A*Ah = B*Bh for all h ∈ Dom(A*A), then there exists a partial isometry U such that A = UB. U is unique if Ran(B)⊥⊂ Ker(U). The operator A being closed and densely defined ensures that the operator A*A is self-adjoint (with dense domain) and therefore allows one to define (A*A)½. Applying the lemma gives polar decomposition.

If an unbounded operator A is affiliated to a von Neumann algebra M, and A = UP is its polar decomposition, then U is in M and so is the spectral projection of P, 1B(P), for any Borel set B in [0, ∞).

[edit] Applications

Continuum mechanics uses polar decomposition to separate deformation into stretching and rotation, where P represents stretching and U the rotation.

[edit] See also

[edit] References

  • Conway, J.B.: A Course in Functional Analysis. Graduate Texts in Mathematics. New York: Springer 1990
  • Douglas, R.G.: On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space. Proc. Amer. Math. Soc. 17, 413-415 (1966)