Open main menu

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 decompositionEdit

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 = WΣV*, 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 SUS* = Φ−1, where Φ is a diagonal unitary matrix of phases e. Putting Q = VS*, 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 existenceEdit

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   normalEdit

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   invertibleEdit

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.

Yet another way to directly show the unitarity of   is to note that, writing the SVD of   in terms of rank-1 matrices as  , where  are the singular values of  , we have

which directly implies the unitarity of   because a matrix is unitary if and only if its singular values have unitary absolute value.

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

General caseEdit

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 proofEdit

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 spaceEdit

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 operatorsEdit

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 decompositionEdit

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 decompositionsEdit

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

  • If x ≠ 0, z = x(1 + ε(y/x)) is a polar decomposition of a dual number z = x + yε, 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 x2y2, then the unit hyperbola x2y2 = 1 and its conjugate x2y2 = −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 j2 = +1 and the arithmetic[6] of split-complex numbers is used. The branch through (−1, 0) is traced by −eaj. Since the operation of multiplying by j reflects a point across the line y = x, the second hyperbola has branches traced by jeaj or −jeaj. 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 decompositionEdit

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's 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 is also applicable to singular or rectangular matrices A.

See alsoEdit


  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 doi:10.1137/070699895. ISSN 0895-4798.