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 polar decomposition of a nonzero complex number z

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

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

1 Matrix polar decomposition
2 Bounded operators on Hilbert space
3 Unbounded operators
4 See also

[edit] Matrix polar decomposition

The polar decomposition of 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 $det P = r = | det A |$ and $det U = e i θ$ .

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

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

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 e^iφ. 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

[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 R(A).

The proof uses the same lemma as above. Notice that the lemma goes through without requiring that A be closed. Closedness ensures that the operator A*A is self-adjoint and therefore allows one to define (A*A)^½. The rest of the argument is same as the bounded case.