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 as where r is the absolute value of z (a positive real number), and is an element of the circle group.

Matrix polar decomposition[edit]

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

where U is a unitary matrix and P is a positive-semidefinite Hermitian matrix.[1] Intuitively, the polar decomposition separates A into a component that stretches the space along a set of orthogonal axes, represented by P, and a rotation (with possible reflection) represented by U. The decomposition of the complex conjugate of is given by

This decomposition always exists; and so long as A is invertible, it is unique, with P positive-definite. Note that

gives the corresponding polar decomposition of the determinant of A, since and . In particular, if has determinant 1 then both and have determinant 1.

The matrix P is always unique, even if A is singular, and given by

where A* denotes the conjugate transpose of A. This expression is ensured to be well-defined, since is a positive-semidefinite Hermitian matrix, and therefore has a unique positive-semidefinite Hermitian square root.[2] If A is invertible, then the matrix U is uniquely determined by

Moreover, if is invertible, then is strictly positive definite, and thus has a unique self-adjoint logarithm. Every invertible matrix can therefore be written uniquely in the form

where is unitary and is self-adjoint.[3] This decomposition is useful in computing the fundamental group of (matrix) Lie groups.[4]

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

confirming that P is positive-definite and U is unitary. Thus, the existence of the SVD is equivalent to the existence of polar decomposition.

One can also decompose A in the form

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

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 e. Putting Q = V S*, one can then re-write the polar decomposition as

so A then thus also has a spectral decomposition

with complex eigenvalues such that and a unitary matrix of complex eigenvectors Q.

The polar decomposition of a square invertible real matrix A is of the form

where is a positive-definite matrix and is an orthogonal matrix.

Construction and proofs of existence[edit]

The core idea behind the construction of the polar decomposition is similar to that used to compute the singular-value decomposition.

For any matrix , the matrix is hermitian and positive semi-definite, and therefore unitarily equivalent to a positive semi-definite diagonal matrix. Let then be the unitary matrix such that , with diagonal and positive semi-definite.

Case of normal[edit]

If is normal, then it is unitarily equivalent to a diagonal matrix: for some unitary and some diagonal matrix .

The polar decomposition is in this case obtained by writing

where is the diagonal matrix with the absolute values of the elements of , and is a diagonal matrix with containing the phases of the elements of . In other words,

When , the corresponding phase can be chosen arbitrarily.

Going back into the original basis, we obtain the polar decomposition of :

Case of invertible[edit]

Using for example the singular-value decomposition, it can be readily shown that a matrix is invertible if and only if (equivalently, ) is. Moreover, this is true if and only if the eigenvalues of are all not zero[5].

In this case, the polar decomposition is directly obtained by writing

and observing that is unitary. To see this, we can exploit the spectral decomposition of to write .

In this expression, is unitary because is. To show that also is unitary, we can use the SVD to write , so that

where again is unitary by construction.

Note how, from the above construction, it follows that the unitary matrix in the polar decomposition of an invertible matrix is uniquely defined.

General case[edit]

The above argument crucially relies on the existence of , and therefore on being invertible. Indeed, in the general case, is not generally well-defined, due to the possibility of having vanishing eigenvalues.

Let us denote with the (in general not square) matrix whose columns are the eigenvectors of corresponding to non-vanishing eigenvalues, with the diagonal matrix containing the associated non-zero eigenvalues, and with the matrix with the remaining eigenvectors of . We can then write the spectral decomposition of as:

Note that, similarly to the invertible case, is well-defined and its columns are orthonormal, although it is not in general square and therefore unitary.

We now define

where is a matrix whose columns are chosen so that is unitary. This is done by finding a set of orthonormal vectors which, together with the columns of , form a complete base for the space, and using these vectors as the columns of . Note how the definition of is not unique, unless (and therefore ) is invertible, in which case is already unitary and uniquely defined.

The argument now proceeds similarly to the invertible case, with the only difference of using in place of . Indeed, we have:

where we used the orthogonality of the columns of and , which is equivalent to , and is the product of two unitaries, and is therefore also unitary.

General case, alternative proof[edit]

Making use of the SVD of , a more direct proof can be found.

The SVD of reads , with unitary matrices, and a diagonal, positive semi-definite matrix. By simply inserting an additional pair of s or s, we obtain the two forms of the polar decomposition of :

Bounded operators on Hilbert space[edit]

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.

The operator U must be weakened to a partial isometry, rather than unitary, because of the following issues. If A is the one-sided shift on l2(N), then |A| = {A*A}½ = I. So if A = U |A|, U must be A, which is not unitary.

The existence of a polar decomposition is a consequence of Douglas' lemma:

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

The operator C can be defined by C(Bh) := Ah for all h in H, extended by continuity to the closure of Ran(B), and by zero on the orthogonal complement to all of H. The lemma then follows since A*AB*B implies Ker(B) ⊂ Ker(A).

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,

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

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 a 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.

Unbounded operators[edit]

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

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 hDom(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, ∞).

Quaternion polar decomposition[edit]

The polar decomposition of quaternions H depends on the sphere of square roots of minus one. Given any r on this sphere, and an angle –π < a ≤ π, the versor is on the 3-sphere of H. For a = 0 and a = π, the versor is 1 or −1 regardless of which r is selected. The norm t of a quaternion q is the Euclidean distance from the origin to q. When a quaternion is not just a real number, then there is a unique polar decomposition

Alternative planar decompositions[edit]

In the Cartesian plane, alternative planar ring decompositions arise as follows:

  • If x ≠ 0, z = x ( 1 + (y/x) ε) is a polar decomposition of a dual number z = x + y ε, where ε2 = 0, i.e. ε is nilpotent. In this polar decomposition, the unit circle has been replaced by the line x = 1, the polar angle by the slope y/x, and the radius x is negative in the left half-plane.
  • If x2 ≠ y2, then the unit hyperbola x2 − y2 = 1 and its conjugate x2 − y2 = −1 can be used to form a polar decomposition based on the branch of the unit hyperbola through (1,0). This branch is parametrized by the hyperbolic angle a and is written
where j 2 = +1 and the arithmetic[6] of split-complex numbers is used. The branch through (−1,0) is traced by −ea j. Since the operation of multiplying by j reflects a point across the line y = x, the second hyperbola has branches traced by jea j or −jea j. Therefore a point in one of the quadrants has a polar decomposition in one of the forms:
The set { 1, −1, j, −j } has products that make it isomorphic to the Klein four-group. Evidently polar decomposition in this case involves an element from that group.

Numerical determination of the matrix polar decomposition[edit]

To compute an approximation of the polar decomposition A=UP, usually the unitary factor U is approximated.[7][8] The iteration is based on Heron's method for the square root of 1 and computes, starting from , the sequence

The combination of inversion and Hermite conjugation is chosen so that in the singular value decomposition, the unitary factors remain the same and the iteration reduces to Heron's method on the singular values.

This basic iteration may be refined to speed up the process:

  • Every step or in regular intervals, the range of the singular values of is estimated and then the matrix is rescaled to to center the singular values around 1. The scaling factor is computed using matrix norms of the matrix and its inverse. Examples of such scale estimates are:
using the row-sum and column-sum matrix norms or
using the Frobenius norm. Including the scale factor, the iteration is now
  • The QR decomposition can be used in a preparation step to reduce a singular matrix A to a smaller regular matrix, and inside every step to speed up the computation of the inverse.
  • Heron' method for computing roots of can be replaced by higher order methods, for instance based on Halley's method of third order, resulting in
This iteration can again be combined with rescaling. This particular formula has the benefit that it also applicable to singular or rectangular matrices A.

See also[edit]

References[edit]

  1. ^ Hall 2015 Section 2.5
  2. ^ Hall 2015 Lemma 2.18
  3. ^ Hall 2015 Theorem 2.17
  4. ^ Hall 2015 Section 13.3
  5. ^ Note how this implies, by the positivity of , that the eigenvalues are all real and strictly positive.
  6. ^ Sobczyk, G.(1995) "Hyperbolic Number Plane", College Mathematics Journal 26:268–80
  7. ^ Higham, Nicholas J. (1986). "Computing the polar decomposition with applications". SIAM J. Sci. Stat. Comput. Philadelphia, PA, USA: Society for Industrial and Applied Mathematics. 7 (4): 1160–1174. doi:10.1137/0907079. ISSN 0196-5204. Archived from the original on 2013-05-08.
  8. ^ Byers, Ralph; Hongguo Xu (2008). "A New Scaling for Newton's Iteration for the Polar Decomposition and its Backward Stability". SIAM J. Matrix Anal. Appl. Philadelphia, PA, USA: Society for Industrial and Applied Mathematics. 30 (2): 822–843. CiteSeerX 10.1.1.378.6737. doi:10.1137/070699895. ISSN 0895-4798.