# Polar decomposition

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 ${\displaystyle z=re^{i\theta }\,}$ where r is the absolute value of z (a positive real number), and ${\displaystyle e^{i\theta }}$ is an element of the circle group.

## Matrix polar decomposition

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

${\displaystyle A=UP}$

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

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

${\displaystyle \det A=\det U\det P=e^{i\theta }\cdot r}$

gives the corresponding polar decomposition of the determinant of A, since ${\displaystyle \det U=e^{i\theta }}$ and ${\displaystyle \det P=r=|\det A|}$. In particular, if ${\displaystyle A}$ has determinant 1 then both ${\displaystyle U}$ and ${\displaystyle P}$ have determinant 1.

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

${\displaystyle P=\left(A^{*}A\right)^{\frac {1}{2}},}$

where A* denotes the conjugate transpose of A. This expression is ensured to be well-defined, since ${\displaystyle A^{*}A}$ 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

${\displaystyle U=AP^{-1}.}$

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

${\displaystyle A=Ue^{X},}$

where ${\displaystyle U}$ is unitary and ${\displaystyle X}$ 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

{\displaystyle {\begin{aligned}P&=V\Sigma V^{*}\\U&=WV^{*}\end{aligned}}}

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

${\displaystyle A=P'U}$

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

${\displaystyle P'=UPU^{-1}=\left(AA^{*}\right)^{\frac {1}{2}}=W\Sigma W^{*}.}$

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 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

${\displaystyle A=\left(Q\Phi Q^{*}\right)\left(Q\Sigma Q^{*}\right),\,}$

so A then thus also has a spectral decomposition

${\displaystyle A=Q\Lambda Q^{*}}$

with complex eigenvalues such that ${\displaystyle \Lambda \Lambda ^{*}=\Sigma ^{2}}$ and a unitary matrix of complex eigenvectors Q.

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

${\displaystyle A=|A|R}$

where ${\displaystyle |A|=\left(AA^{\textsf {T}}\right)^{\frac {1}{2}}}$ is a positive-definite matrix and ${\displaystyle R=|A|^{-1}A}$ is an orthogonal matrix.

### Construction and proofs of existence

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

For any matrix ${\displaystyle A}$, the matrix ${\displaystyle A^{*}A}$ is hermitian and positive semi-definite, and therefore unitarily equivalent to a positive semi-definite diagonal matrix. Let then ${\displaystyle V}$ be the unitary matrix such that ${\displaystyle A^{*}A=VDV^{*}}$, with ${\displaystyle D}$ diagonal and positive semi-definite.

#### Case of ${\displaystyle A}$ normal

If ${\displaystyle A}$ is normal, then it is unitarily equivalent to a diagonal matrix: ${\displaystyle A=V\Lambda V^{*}}$ for some unitary ${\displaystyle V}$ and some diagonal matrix ${\displaystyle \Lambda }$.

The polar decomposition is in this case obtained by writing

${\displaystyle V^{*}AV=\Phi _{\Lambda }|\Lambda |,}$

where ${\displaystyle |\Lambda |}$ is the diagonal matrix with the absolute values of the elements of ${\displaystyle \Lambda }$, and ${\displaystyle \Phi _{\Lambda }}$ is a diagonal matrix with containing the phases of the elements of ${\displaystyle \Lambda }$. In other words,

${\displaystyle \left(\Phi _{\Lambda }|\Lambda |\right)_{jk}=\delta _{jk}e^{i\phi _{k}}\left|\lambda _{k}\right|=\delta _{jk}\lambda _{k}.}$

When ${\displaystyle \lambda _{k}=0}$, the corresponding phase can be chosen arbitrarily.

Going back into the original basis, we obtain the polar decomposition of ${\displaystyle A}$:

${\displaystyle A=V\Phi _{\Lambda }|\Lambda |V^{*}=\underbrace {\left(V\Phi _{\Lambda }V^{*}\right)} _{U}\underbrace {\left(V|\Lambda |V^{*}\right)} _{P}.}$

#### Case of ${\displaystyle A}$ invertible

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

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

${\displaystyle A=A\left(A^{*}A\right)^{-{\frac {1}{2}}}\left(A^{*}A\right)^{\frac {1}{2}},}$

and observing that ${\displaystyle A\left(A^{*}A\right)^{-{\frac {1}{2}}}}$ is unitary. To see this, we can exploit the spectral decomposition of ${\displaystyle A^{*}A}$ to write ${\displaystyle A\left(A^{*}A\right)^{-{\frac {1}{2}}}=AVD^{-{\frac {1}{2}}}V^{*}}$.

In this expression, ${\displaystyle V^{*}}$ is unitary because ${\displaystyle V}$ is. To show that also ${\displaystyle AVD^{-{\frac {1}{2}}}}$ is unitary, we can use the SVD to write ${\displaystyle A=WD^{\frac {1}{2}}V^{*}}$, so that

${\displaystyle AVD^{-{\frac {1}{2}}}=WD^{\frac {1}{2}}V^{*}VD^{-{\frac {1}{2}}}=W,}$

where again ${\displaystyle W}$ is unitary by construction.

Yet another way to directly show the unitarity of ${\displaystyle A\left(A^{*}A\right)^{-{\frac {1}{2}}}}$ is to note that, writing the SVD of ${\displaystyle A}$ in terms of rank-1 matrices as ${\displaystyle A=\sum _{k}s_{k}v_{k}w_{k}^{*}}$, where ${\displaystyle s_{k}}$are the singular values of ${\displaystyle A}$, we have

${\displaystyle A\left(A^{*}A\right)^{-{\frac {1}{2}}}=\left(\sum _{j}\lambda _{j}v_{j}w_{j}^{*}\right)\left(\sum _{k}|\lambda _{k}|^{-1}w_{k}w_{k}^{*}\right)=\sum _{k}{\frac {\lambda _{k}}{|\lambda _{k}|}}v_{k}w_{k}^{*},}$
which directly implies the unitarity of ${\displaystyle A\left(A^{*}A\right)^{-{\frac {1}{2}}}}$ 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 case

The above argument crucially relies on the existence of ${\displaystyle \left(A^{*}A\right)^{-{\frac {1}{2}}}}$, and therefore on ${\displaystyle A^{*}A}$ being invertible. Indeed, in the general case, ${\displaystyle AVD^{-{\frac {1}{2}}}}$ is not generally well-defined, due to the possibility of ${\displaystyle D}$ having vanishing eigenvalues.

Let us denote with ${\displaystyle V_{1}}$ the (in general not square) matrix whose columns are the eigenvectors of ${\displaystyle A^{*}A}$ corresponding to non-vanishing eigenvalues, with ${\displaystyle D_{1}}$ the diagonal matrix containing the associated non-zero eigenvalues, and with ${\displaystyle V_{2}}$ the matrix with the remaining eigenvectors of ${\displaystyle A^{*}A}$. We can then write the spectral decomposition of ${\displaystyle A^{*}A}$ as:

${\displaystyle A^{*}A={\begin{bmatrix}V_{1}&V_{2}\end{bmatrix}}{\begin{bmatrix}D_{1}&0\\0&0\end{bmatrix}}{\begin{bmatrix}V_{1}^{*}\\V_{2}^{*}\end{bmatrix}}=V_{1}D_{1}V_{1}^{*}.}$

Note that, similarly to the invertible case, ${\displaystyle AV_{1}D_{1}^{-{\frac {1}{2}}}}$ is well-defined and its columns are orthonormal, although it is not in general square and therefore unitary.

We now define

${\displaystyle U'=\left[AV_{1}D_{1}^{-{\frac {1}{2}}},\Phi \right],}$

where ${\displaystyle \Phi }$ is a matrix whose columns are chosen so that ${\displaystyle U'}$ is unitary. This is done by finding a set of orthonormal vectors which, together with the columns of ${\displaystyle AV_{1}D_{1}^{-{\frac {1}{2}}}}$, form a complete base for the space, and using these vectors as the columns of ${\displaystyle \Phi }$. Note how the definition of ${\displaystyle U'}$ is not unique, unless ${\displaystyle A^{*}A}$ (and therefore ${\displaystyle A}$) is invertible, in which case ${\displaystyle AV_{1}D_{1}^{-{\frac {1}{2}}}}$ is already unitary and uniquely defined.

The argument now proceeds similarly to the invertible case, with the only difference of using ${\displaystyle U'}$ in place of ${\displaystyle AVD^{-{\frac {1}{2}}}}$. Indeed, we have:

${\displaystyle U\left(A^{*}A\right)^{\frac {1}{2}}\equiv U'{\begin{bmatrix}V_{1}^{*}\\V_{2}^{*}\end{bmatrix}}\left(A^{*}A\right)^{\frac {1}{2}}=\left(AV_{1}D_{1}^{-{\frac {1}{2}}}V_{1}^{*}+\Phi V_{2}^{*}\right)V_{1}D_{1}^{\frac {1}{2}}V_{1}=A,}$

where we used the orthogonality of the columns of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$, which is equivalent to ${\displaystyle V_{2}^{*}V_{1}=0}$, and ${\displaystyle U}$ is the product of two unitaries, and is therefore also unitary.

#### General case, alternative proof

Making use of the SVD of ${\displaystyle A}$, a more direct proof can be found.

The SVD of ${\displaystyle A}$ reads ${\displaystyle A=WD^{\frac {1}{2}}V^{*}}$, with ${\displaystyle W,V}$ unitary matrices, and ${\displaystyle D}$ a diagonal, positive semi-definite matrix. By simply inserting an additional pair of ${\displaystyle W}$s or ${\displaystyle V}$s, we obtain the two forms of the polar decomposition of ${\displaystyle A}$:

${\displaystyle A=WD^{\frac {1}{2}}V^{*}=\underbrace {\left(WD^{\frac {1}{2}}W^{*}\right)} _{P}\underbrace {\left(WV^{*}\right)} _{U}=\underbrace {\left(WV^{*}\right)} _{U}\underbrace {\left(VD^{\frac {1}{2}}V^{*}\right)} _{P'}.}$

## 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.

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,

${\displaystyle A^{*}A=\left(A^{*}A\right)^{\frac {1}{2}}\left(A^{*}A\right)^{\frac {1}{2}},}$

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

${\displaystyle A=U\left(A^{*}A\right)^{\frac {1}{2}}}$

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 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

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

${\displaystyle A=U|A|\,}$

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 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

The polar decomposition of quaternions H depends on the sphere ${\displaystyle \lbrace xi+yj+zk\in H:x^{2}+y^{2}+z^{2}=1\rbrace }$ of square roots of minus one. Given any r on this sphere, and an angle −π < a ≤ π, the versor ${\displaystyle e^{ar}=\cos(a)+r\ \sin(a)}$ 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 ${\displaystyle q=te^{ar}.}$

## Alternative planar decompositions

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
${\displaystyle \cosh(a)+j\ \sinh(a)=\exp(aj)=e^{aj}}$

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:

${\displaystyle re^{aj},-re^{aj},rje^{aj},-rje^{aj},\quad r>0}$
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

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 ${\displaystyle U_{0}=A}$, the sequence

${\displaystyle U_{k+1}={\frac {1}{2}}\left(U_{k}+\left(U_{k}^{*}\right)^{-1}\right),\qquad k=0,1,2,\ldots }$

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 ${\displaystyle U_{k}}$ is estimated and then the matrix is rescaled to ${\displaystyle \gamma _{k}U_{k}}$ to center the singular values around 1. The scaling factor ${\displaystyle \gamma _{k}}$ is computed using matrix norms of the matrix and its inverse. Examples of such scale estimates are:
${\displaystyle \gamma _{k}={\sqrt[{4}]{\frac {\left\|U_{k}^{-1}\right\|_{1}\left\|U_{k}^{-1}\right\|_{\infty }}{\left\|U_{k}\right\|_{1}\left\|U_{k}\right\|_{\infty }}}}}$

using the row-sum and column-sum matrix norms or

${\displaystyle \gamma _{k}={\sqrt {\frac {\left\|U_{k}^{-1}\right\|_{F}}{\left\|U_{k}\right\|_{F}}}}}$

using the Frobenius norm. Including the scale factor, the iteration is now

${\displaystyle U_{k+1}={\frac {1}{2}}\left(\gamma _{k}U_{k}+{\frac {1}{\gamma _{k}}}\left(U_{k}^{*}\right)^{-1}\right),\qquad k=0,1,2,\ldots }$
• 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 ${\displaystyle x^{2}-1=0}$ can be replaced by higher order methods, for instance based on Halley's method of third order, resulting in
${\displaystyle U_{k+1}=U_{k}\left(I+3U_{k}^{*}U_{k}\right)^{-1}\left(3I+U_{k}^{*}U_{k}\right),\qquad k=0,1,2,\ldots }$
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.

5. ^ Note how this implies, by the positivity of ${\displaystyle A^{*}A}$, that the eigenvalues are all real and strictly positive.