In mathematics, the polar decomposition of a square real or complex matrix is a factorization of the form , where is a unitary matrix and is a positive semidefinite Hermitian matrix ( is an orthogonal matrix and is a positive semidefinite symmetric matrix in the real case), both square and of the same size.^{[1]}
If a real matrix is interpreted as a linear transformation of dimensional space , the polar decomposition separates it into a rotation or reflection of , and a scaling of the space along a set of orthogonal axes.
The polar decomposition of a square matrix always exists. If is invertible, the decomposition is unique, and the factor will be positivedefinite. In that case, can be written uniquely in the form , where is unitary and is the unique selfadjoint logarithm of the matrix .^{[2]} This decomposition is useful in computing the fundamental group of (matrix) Lie groups.^{[3]}
The polar decomposition can also be defined as where is a symmetric positivedefinite matrix with the same eigenvalues as but different eigenvectors.
The polar decomposition of a matrix can be seen as the matrix analog of the polar form of a complex number as , where is its absolute value (a nonnegative real number), and is a complex number with unit norm (an element of the circle group).
The definition may be extended to rectangular matrices by requiring to be a semiunitary matrix and to be a positivesemidefinite Hermitian matrix. The decomposition always exists and is always unique. The matrix is unique if and only if has full rank. ^{[4]}
A real square matrix can be interpreted as the linear transformation of that takes a column vector to . Then, in the polar decomposition , the factor is an real orthonormal matrix. The polar decomposition then can be seen as expressing the linear transformation defined by into a scaling of the space along each eigenvector of by a scale factor (the action of ), followed by a rotation of (the action of ).
Alternatively, the decomposition expresses the transformation defined by as a rotation () followed by a scaling () along certain orthogonal directions. The scale factors are the same, but the directions are different.
The polar decomposition of the complex conjugate of is given by Note thatgives the corresponding polar decomposition of the determinant of A, since and . In particular, if has determinant 1 then both and have determinant 1.
The positivesemidefinite matrix P is always unique, even if A is singular, and is denoted aswhere denotes the conjugate transpose of . The uniqueness of P ensures that this expression is welldefined. The uniqueness is guaranteed by the fact that is a positivesemidefinite Hermitian matrix and, therefore, has a unique positivesemidefinite Hermitian square root.^{[5]} If A is invertible, then P is positivedefinite, thus also invertible and the matrix U is uniquely determined by
In terms of the singular value decomposition (SVD) of , , one haswhere , , and are unitary matrices (called orthogonal matrices if the field is the reals ). This confirms that is positivedefinite and is unitary. Thus, the existence of the SVD is equivalent to the existence of polar decomposition.
One can also decompose in the formHere is the same as before and is given byThis 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 polar decomposition of a square invertible real matrix is of the form where is a positivedefinite matrix and is an orthogonal matrix.
The matrix with polar decomposition is normal if and only if and commute: , or equivalently, they are simultaneously diagonalizable.
The core idea behind the construction of the polar decomposition is similar to that used to compute the singularvalue decomposition.
If is normal, then it is unitarily equivalent to a diagonal matrix: for some unitary matrix and some diagonal matrix . This makes the derivation of its polar decomposition particularly straightforward, as we can then write where is a diagonal matrix containing the phases of the elements of , that is, when , and when .
The polar decomposition is thus , with and diagonal in the eigenbasis of and having eigenvalues equal to the phases and absolute values of those of , respectively.
From the singularvalue decomposition, it can be 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.^{[6]}
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 rank1 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.
The SVD of a square matrix reads , with unitary matrices, and a diagonal, positive semidefinite matrix. By simply inserting an additional pair of s or s, we obtain the two forms of the polar decomposition of :More generally, if is some rectangular matrix, its SVD can be written as where now and are isometries with dimensions and , respectively, where , and is again a diagonal positive semidefinite square matrix with dimensions . We can now apply the same reasoning used in the above equation to write , but now is not in general unitary. Nonetheless, has the same support and range as , and it satisfies and . This makes into an isometry when its action is restricted onto the support of , that is, it means that is a partial isometry.
As an explicit example of this more general case, consider the SVD of the following matrix:We then havewhich is an isometry, but not unitary. On the other hand, if we consider the decomposition ofwe findwhich is a partial isometry (but not an isometry).
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 nonnegative 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 nonnegative selfadjoint 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 onesided shift on l^{2}(N), then A = {A^{*}A}^{1/2} = 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^{*}A ≤ B^{*}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^{*}A ≤ B^{*}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)^{1/2} 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)^{1/2} 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 finitedimensional, 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.
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) nonnegative 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 selfadjoint (with dense domain) and therefore allows one to define (A^{*}A)^{1/2}. 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, 1_{B}(P), for any Borel set B in [0, ∞).
The polar decomposition of quaternions with orthonormal basis quaternions depends on the unit 2dimensional sphere of square roots of minus one, known as right versors. Given any on this sphere, and an angle −π < a ≤ π , the versor is on the unit 3sphere of 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:
Here r, a, t are all uniquely determined such that r is a right versor ( r ^{2} = –1 ), a satisfies 0 < a < π , and t > 0 .
In the Cartesian plane, alternative planar ring decompositions arise as follows:
where j^{2} = +1 and the arithmetic^{[7]} of splitcomplex numbers is used. The branch through (−1, 0) is traced by −e^{aj}. Since the operation of multiplying by j reflects a point across the line y = x, the conjugate hyperbola has branches traced by je^{aj} or −je^{aj}. 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 fourgroup. Evidently polar decomposition in this case involves an element from that group.To compute an approximation of the polar decomposition A = UP, usually the unitary factor U is approximated.^{[8]}^{[9]} 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:
using the rowsum and columnsum matrix norms or using the Frobenius norm. Including the scale factor, the iteration is now
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