Polar decomposition

A real square ${\displaystyle m\times m}$  matrix ${\displaystyle A}$  can be interpreted as the linear transformationof${\displaystyle \mathbb {R} ^{m}}$  that takes a column vector ${\displaystyle x}$ to${\displaystyle Ax}$ . Then, in the polar decomposition ${\displaystyle A=RP}$ , the factor ${\displaystyle R}$  is an ${\displaystyle m\times m}$  real orthonormal matrix. The polar decomposition then can be seen as expressing the linear transformation defined by ${\displaystyle A}$  into a scaling of the space ${\displaystyle \mathbb {R} ^{m}}$  along each eigenvector ${\displaystyle e_{i}}$ of${\displaystyle P}$  by a scale factor ${\displaystyle \sigma _{i}}$  (the action of ${\displaystyle P}$ ), followed by a rotation of ${\displaystyle \mathbb {R} ^{m}}$  (the action of ${\displaystyle R}$ ).

Alternatively, the decomposition ${\displaystyle A=PR}$  expresses the transformation defined by ${\displaystyle A}$  as a rotation (${\displaystyle R}$ ) followed by a scaling (${\displaystyle P}$ ) along certain orthogonal directions. The scale factors are the same, but the directions are different.

The polar decomposition of the complex conjugateof${\displaystyle A}$  is given by ${\displaystyle {\overline {A}}={\overline {U}}{\overline {P}}.}$  Note that$${\displaystyle \det A=\det U\det P=e^{i\theta }r}$$ gives the corresponding polar decomposition of the determinantofA, since ${\displaystyle \det U=e^{i\theta }}$  and ${\displaystyle \det P=r=\left|\det A\right|}$ . In particular, if ${\displaystyle A}$  has determinant 1 then both ${\displaystyle U}$  and ${\displaystyle P}$  have determinant 1.

The positive-semidefinite matrix P is always unique, even if Aissingular, and is denoted as$${\displaystyle P=\left(A^{*}A\right)^{1/2},}$$ where ${\displaystyle A^{*}}$  denotes the conjugate transposeof${\displaystyle A}$ . The uniqueness of P ensures that this expression is well-defined. The uniqueness is guaranteed by the fact that ${\displaystyle A^{*}A}$  is a positive-semidefinite Hermitian matrix and, therefore, has a unique positive-semidefinite Hermitian square root.^[5]IfA is invertible, then P is positive-definite, thus also invertible and the matrix U is uniquely determined by$${\displaystyle U=AP^{-1}.}$$ 

In terms of the singular value decomposition (SVD) of ${\displaystyle A}$ , ${\displaystyle A=W\Sigma V^{*}}$ , one has$${\displaystyle {\begin{aligned}P&=V\Sigma V^{*}\\U&=WV^{*}\end{aligned}}}$$ where ${\displaystyle U}$ , ${\displaystyle V}$ , and ${\displaystyle W}$  are unitary matrices (called orthogonal matrices if the field is the reals ${\displaystyle \mathbb {R} }$ ). This confirms that ${\displaystyle P}$  is positive-definite and ${\displaystyle U}$  is unitary. Thus, the existence of the SVD is equivalent to the existence of polar decomposition.

One can also decompose ${\displaystyle A}$  in the form$${\displaystyle A=P'U}$$ Here ${\displaystyle U}$  is the same as before and ${\displaystyle P'}$  is given by$${\displaystyle P'=UPU^{-1}=\left(AA^{*}\right)^{1/2}=W\Sigma W^{*}.}$$ 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 polar decomposition of a square invertible real matrix ${\displaystyle A}$  is of the form $${\displaystyle A=|A|R}$$  where ${\displaystyle |A|=\left(AA^{\textsf {T}}\right)^{1/2}}$  is a positive-definite matrix and ${\displaystyle R=|A|^{-1}A}$  is an orthogonal matrix.

The matrix ${\displaystyle A}$  with polar decomposition ${\displaystyle A=UP}$ isnormal if and only if ${\displaystyle U}$  and ${\displaystyle P}$  commute: ${\displaystyle UP=PU}$ , or equivalently, they are simultaneously diagonalizable.

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

If${\displaystyle A}$ isnormal, then it is unitarily equivalent to a diagonal matrix: ${\displaystyle A=V\Lambda V^{*}}$  for some unitary matrix ${\displaystyle V}$  and some diagonal matrix ${\displaystyle \Lambda }$ . This makes the derivation of its polar decomposition particularly straightforward, as we can then write $${\displaystyle A=V\Phi _{\Lambda }|\Lambda |V^{*}=\underbrace {\left(V\Phi _{\Lambda }V^{*}\right)} _{\equiv U}\underbrace {\left(V|\Lambda |V^{*}\right)} _{\equiv P},}$$  where ${\displaystyle \Phi _{\Lambda }}$  is a diagonal matrix containing the phases of the elements of ${\displaystyle \Lambda }$ , that is, ${\displaystyle (\Phi _{\Lambda })_{ii}\equiv \Lambda _{ii}/|\Lambda _{ii}|}$  when ${\displaystyle \Lambda _{ii}\neq 0}$ , and ${\displaystyle (\Phi _{\Lambda })_{ii}=0}$  when ${\displaystyle \Lambda _{ii}=0}$ .

The polar decomposition is thus ${\displaystyle A=UP}$ , with ${\displaystyle U}$  and ${\displaystyle P}$  diagonal in the eigenbasis of ${\displaystyle A}$  and having eigenvalues equal to the phases and absolute values of those of ${\displaystyle A}$ , respectively.

From the singular-value decomposition, it can be 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.^[6]

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

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

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

The SVD of a square matrix ${\displaystyle A}$  reads ${\displaystyle A=WD^{1/2}V^{*}}$ , 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 ${\displaystyle A}$ :$${\displaystyle A=WD^{1/2}V^{*}=\underbrace {\left(WD^{1/2}W^{*}\right)} _{P}\underbrace {\left(WV^{*}\right)} _{U}=\underbrace {\left(WV^{*}\right)} _{U}\underbrace {\left(VD^{1/2}V^{*}\right)} _{P'}.}$$ More generally, if ${\displaystyle A}$  is some rectangular ${\displaystyle n\times m}$  matrix, its SVD can be written as ${\displaystyle A=WD^{1/2}V^{*}}$  where now ${\displaystyle W}$  and ${\displaystyle V}$  are isometries with dimensions ${\displaystyle n\times r}$  and ${\displaystyle m\times r}$ , respectively, where ${\displaystyle r\equiv \operatorname {rank} ($A)}  , and ${\displaystyle D}$  is again a diagonal positive semi-definite square matrix with dimensions ${\displaystyle r\times r}$ . We can now apply the same reasoning used in the above equation to write ${\displaystyle A=PU=UP'}$ , but now ${\displaystyle U\equiv WV^{*}}$  is not in general unitary. Nonetheless, ${\displaystyle U}$  has the same support and range as ${\displaystyle A}$ , and it satisfies ${\displaystyle U^{*}U=VV^{*}}$  and ${\displaystyle UU^{*}=WW^{*}}$ . This makes ${\displaystyle U}$  into an isometry when its action is restricted onto the support of ${\displaystyle A}$ , that is, it means that ${\displaystyle U}$  is a partial isometry.

As an explicit example of this more general case, consider the SVD of the following matrix:$${\displaystyle A\equiv {\begin{pmatrix}1&1\\2&-2\\0&0\end{pmatrix}}=\underbrace {\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}} _{\equiv W}\underbrace {\begin{pmatrix}{\sqrt {2}}&0\\0&{\sqrt {8}}\end{pmatrix}} _{\sqrt {D}}\underbrace {\begin{pmatrix}{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\\{\frac {1}{\sqrt {2}}}&-{\frac {1}{\sqrt {2}}}\end{pmatrix}} _{V^{\dagger }}.}$$ We then have$${\displaystyle WV^{\dagger }={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&1\\1&-1\\0&0\end{pmatrix}}}$$ which is an isometry, but not unitary. On the other hand, if we consider the decomposition of$${\displaystyle A\equiv {\begin{pmatrix}1&0&0\\0&2&0\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}{\begin{pmatrix}1&0\\0&2\end{pmatrix}}{\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}},}$$ we find$${\displaystyle WV^{\dagger }={\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}},}$$ which 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 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 shiftonl²(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:

The operator C can be defined by C(Bh) := Ah for all hinH, 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, $${\displaystyle A^{*}A=\left(A^{*}A\right)^{1/2}\left(A^{*}A\right)^{1/2},}$$  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 $${\displaystyle A=U\left(A^{*}A\right)^{1/2}}$$  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 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.

IfA 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 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 self-adjoint (with dense domain) and therefore allows one to define (A^*A)^1/2. Applying the lemma gives polar decomposition.

If an unbounded operator Aisaffiliated 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 Bin[0, ∞).

The polar decomposition of quaternions ${\displaystyle \ \mathbb {H} \ }$  with orthonormal basis quaternions ${\displaystyle \ 1,{\widehat {i}},{\widehat {j}},{\widehat {k}}\ }$  depends on the unit 2-dimensional sphere ${\displaystyle \ {\widehat {r}}\in \lbrace \ x\ {\widehat {i}}+y\ {\widehat {j}}+z\ {\widehat {k}}\in \mathbb {H} \smallsetminus \mathbb {R} \ :\ x^{2}+y^{2}+z^{2}=1\ \rbrace \ }$ ofsquare roots of minus one, known as right versors. Given any ${\displaystyle \ {\widehat {r}}\ }$  on this sphere, and an angle −π < a ≤ π , the versor ${\displaystyle \ e^{a\ {\widehat {r}}}{=}\cos($a)+{\widehat {r}}\ \sin(a)\ }   is on the unit 3-sphereof${\displaystyle \ \mathbb {H} ~.}$  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:

${\displaystyle \ q=t\ \exp \left(\ a\ {\widehat {r}}\ \right)~.}$ 

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:

• Ifx ≠ 0, z = x(1 + ε(y/x)) is a polar decomposition of a dual number z = x + yε, where ε² = 0; i.e., εisnilpotent. 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.
• Ifx² ≠ 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

  $${\displaystyle \cosh($$a)+j\ \sinh(a)=\exp(aj)=e^{aj}}   where j² = +1 and the arithmetic^[7]ofsplit-complex 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: $${\displaystyle re^{aj},-re^{aj},rje^{aj},-rje^{aj},\quad r>0}$$ 

  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.

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 ${\displaystyle U_{0}=A}$ , 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 normsor$${\displaystyle \gamma _{k}={\sqrt {\frac {\left\|U_{k}^{-1}\right\|_{F}}{\left\|U_{k}\right\|_{F}}}}}$$  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.

• Cartan decomposition
• Algebraic polar decomposition
• Polar decomposition of a complex measure
• Lie group decomposition

• Conway, J.B.: A Course in Functional Analysis. Graduate Texts in Mathematics. New York: Springer 1990
• Douglas, R.G.: On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space. Proc. Amer. Math. Soc. 17, 413–415 (1966)
• Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, vol. 222 (2nd ed.), Springer, ISBN 978-3319134666.
• Helgason, Sigurdur (1978), Differential geometry, Lie groups, and symmetric spaces, Academic Press, ISBN 0-8218-2848-7