Inmathematics, thepolar decomposition of a squarereal orcomplexmatrix${\displaystyle A}$ is afactorization of the form${\displaystyle A=UP}$, where${\displaystyle U}$ is aunitary matrix, and${\displaystyle P}$ is apositive semi-definiteHermitian matrix (${\displaystyle U}$ is anorthogonal matrix, and${\displaystyle P}$ is a positive semi-definitesymmetric matrix in the real case), both square and of the same size.^[1]

If a real${\displaystyle n\times n}$ matrix${\displaystyle A}$ is interpreted as alinear transformation of${\displaystyle n}$-dimensionalspace${\displaystyle {\mathbb{R}}^n}$, the polar decomposition separates it into arotation orreflection${\displaystyle U}$ of${\displaystyle {\mathbb{R}}^n}$ and ascaling of the space along a set of${\displaystyle n}$ orthogonal axes.

The polar decomposition of a square matrix${\displaystyle A}$ always exists. If${\displaystyle A}$ isinvertible, the decomposition is unique, and the factor${\displaystyle P}$ will bepositive-definite. In that case,${\displaystyle A}$ can be written uniquely in the form${\displaystyle A=U{e}^X}$, where${\displaystyle U}$ is unitary, and${\displaystyle X}$ is the unique self-adjointlogarithm of the matrix${\displaystyle P}$.^[2] This decomposition is useful in computing thefundamental group of (matrix)Lie groups.^[3]

The polar decomposition can also be defined as${\displaystyle A=P^{\prime }U}$, where${\displaystyle P^{\prime }=UP{U}^{-1}}$ is a symmetric positive-definite matrix with the same eigenvalues as${\displaystyle P}$ but different eigenvectors.

The polar decomposition of a matrix can be seen as the matrix analog of thepolar form of acomplex number${\displaystyle z}$ as${\displaystyle z=ur}$, where${\displaystyle r}$ is itsabsolute value (a non-negativereal number), and${\displaystyle u}$ is a complex number with unit norm (an element of thecircle group).

The definition${\displaystyle A=UP}$ may be extended to rectangular matrices${\displaystyle A\in {\mathbb{C}}^{m\times n}}$ by requiring${\displaystyle U\in {\mathbb{C}}^{m\times n}}$ to be asemi-unitary matrix, and${\displaystyle P\in {\mathbb{C}}^{n\times n}}$ to be a positive-semidefinite Hermitian matrix. The decomposition always exists, and${\displaystyle P}$ is always unique. The matrix${\displaystyle U}$ is unique if and only if${\displaystyle A}$ has full rank.^[4]

A real square${\displaystyle m\times m}$ matrix${\displaystyle A}$ can be interpreted as thelinear transformation of${\displaystyle {\mathbb{R}}^m}$ 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 orthogonal matrix. The polar decomposition then can be seen as expressing the linear transformation defined by${\displaystyle A}$ into ascaling 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 thecomplex conjugate of${\displaystyle A}$ is given by${\displaystyle \overline{A}=\overline{U}\overline{P}.}$ Note that$${\displaystyle detA=detUdetP=e^{i\theta }r}$$gives the corresponding polar decomposition of thedeterminant ofA, since${\displaystyle detU=e^{i\theta },}$ and${\displaystyle detP=r=|detA|.}$ In particular, if${\displaystyle A}$ has determinant 1, then both${\displaystyle U}$ and${\displaystyle P}$ have determinant 1.

The positive-semidefinite matrixP is always unique, even ifA issingular, and is denoted as$${\displaystyle P=(A^{\ast }A{)}^{1/2},}$$where${\displaystyle A^{\ast }}$ denotes theconjugate transpose of${\displaystyle A}$. The uniqueness ofP ensures that this expression is well-defined. The uniqueness is guaranteed by the fact that${\displaystyle A^{\ast }A}$ is a positive-semidefinite Hermitian matrix and, therefore, has a unique positive-semidefinite Hermitiansquare root.^[5] IfA is invertible, thenP is positive-definite, thus also invertible, and the matrixU is uniquely determined by$${\displaystyle U=A{P}^{-1}.}$$

In terms of thesingular value decomposition (SVD) of${\displaystyle A}$,${\displaystyle A=W\mathrm{\Sigma }{V}^{\ast }}$, one haswhere${\displaystyle U}$,${\displaystyle V}$, and${\displaystyle W}$ are unitary matrices (orthogonal 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$${\displaystyle A=P^{\prime }U.}$$Here${\displaystyle U}$ is the same as before, and${\displaystyle P^{\prime }}$ is given by$${\displaystyle P^{\prime }=UP{U}^{-1}=(A{A}^{\ast }{)}^{1/2}=W\mathrm{\Sigma }{W}^{\ast }.}$$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.

Thepolar decomposition of a square invertible real matrix${\displaystyle A}$ is of the form$${\displaystyle A=[A]R,}$$where${\displaystyle [A]\equiv {\left(A{A}^{\mathsf{T}}\right)}^{1/2}}$ is apositive-definitematrix, 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 aresimultaneously diagonalizable.

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

If${\displaystyle A}$ isnormal, then it is unitarily equivalent to a diagonal matrix:${\displaystyle A=V\mathrm{\Lambda }{V}^{\ast }}$ for some unitary matrix${\displaystyle V}$ and some diagonal matrix${\displaystyle \mathrm{\Lambda }.}$ This makes the derivation of its polar decomposition particularly straightforward, as we can then write$${\displaystyle A=V{\mathrm{\Phi }}_{\mathrm{\Lambda }}|\mathrm{\Lambda }|V^{\ast }=\underset{\equiv U}{\underbrace{\left(V{\mathrm{\Phi }}_{\mathrm{\Lambda }}{V}^{\ast }\right)}}\underset{\equiv P}{\underbrace{\left(V|\mathrm{\Lambda }|V^{\ast }\right)}},}$$

where${\displaystyle |\mathrm{\Lambda }|}$ is the matrix of absolute diagonal values, and${\displaystyle {\mathrm{\Phi }}_{\mathrm{\Lambda }}}$ is a diagonal matrix containing thephases of the elements of${\displaystyle \mathrm{\Lambda },}$ that is,${\displaystyle ({\mathrm{\Phi }}_{\mathrm{\Lambda }}{)}_{ii}\equiv {\mathrm{\Lambda }}_{ii}/|{\mathrm{\Lambda }}_{ii}|}$ when${\displaystyle {\mathrm{\Lambda }}_{ii}\ne 0,}$, and${\displaystyle ({\mathrm{\Phi }}_{\mathrm{\Lambda }}{)}_{ii}=0}$ when${\displaystyle {\mathrm{\Lambda }}_{ii}=0.}$

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 thesingular-value decomposition, it can be shown that a matrix${\displaystyle A}$ is invertible if and only if${\displaystyle A^{\ast }A}$ (equivalently,${\displaystyle A{A}^{\ast }}$) is. Moreover, this is true if and only if the eigenvalues of${\displaystyle A^{\ast }A}$ are all not zero.^[6]

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

In this expression,${\displaystyle V^{\ast }}$ is unitary because${\displaystyle V}$ is. To show that also${\displaystyle AV{D}^{-1/2}}$ is unitary, we can use theSVD to write${\displaystyle A=W{D}^{1/2}{V}^{\ast }}$, so that$${\displaystyle AV{D}^{-1/2}=W{D}^{1/2}{V}^{\ast }V{D}^{-1/2}=W,}$$where again${\displaystyle W}$ is unitary by construction.

Yet another way to directly show the unitarity of${\displaystyle A{\left(A^{\ast }A\right)}^{-1/2}}$ is to note that, writing theSVD of${\displaystyle A}$ in terms of rank-1 matrices as$A=\sum _k{s}_k{v}_k{w}_k^{\ast }$, where${\displaystyle s_k}$are the singular values of${\displaystyle A}$, we have$${\displaystyle A{\left(A^{\ast }A\right)}^{-1/2}=\left(\sum _j{\lambda }_j{v}_j{w}_j^{\ast }\right)\left(\sum _k|{\lambda }_k{|}^{-1}{w}_k{w}_k^{\ast }\right)=\sum _k\frac{{\lambda }_k}{|{\lambda }_k|}{v}_k{w}_k^{\ast },}$$ which directly implies the unitarity of${\displaystyle A{\left(A^{\ast }A\right)}^{-1/2}}$ because a matrix is unitary if and only if its singular values have unitary absolute value.

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

The SVD of a square matrix${\displaystyle A}$ reads${\displaystyle A=W{D}^{1/2}{V}^{\ast }}$, 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${\displaystyle A}$:$${\displaystyle A=W{D}^{1/2}{V}^{\ast }=\underset{P}{\underbrace{\left(W{D}^{1/2}{W}^{\ast }\right)}}\underset{U}{\underbrace{\left(W{V}^{\ast }\right)}}=\underset{U}{\underbrace{\left(W{V}^{\ast }\right)}}\underset{P^{\prime }}{\underbrace{\left(V{D}^{1/2}{V}^{\ast }\right)}}.}$$More generally, if${\displaystyle A}$ is some rectangular${\displaystyle n\times m}$ matrix, its SVD can be written as${\displaystyle A=W{D}^{1/2}{V}^{\ast }}$ where now${\displaystyle W}$ and${\displaystyle V}$ are isometries with dimensions${\displaystyle n\times r}$ and${\displaystyle m\times r}$, respectively, where${\displaystyle r\equiv \mathrm{rank}(A)}$, 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${\displaystyle A=PU=U{P}^{\prime }}$, but now${\displaystyle U\equiv W{V}^{\ast }}$ is not in general unitary. Nonetheless,${\displaystyle U}$ has the same support and range as${\displaystyle A}$, and it satisfies${\displaystyle U^{\ast }U=V{V}^{\ast }}$ and${\displaystyle U{U}^{\ast }=W{W}^{\ast }}$. 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 apartial 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).

Thepolar decomposition of anybounded linear operatorA between complexHilbert spaces is a canonical factorization as the product of apartial isometry and a non-negative operator.

The polar decomposition for matrices generalizes as follows: ifA is a bounded linear operator then there is a unique factorization ofA as a productA =UP whereU is a partial isometry,P is a non-negative self-adjoint operator and the initial space ofU is the closure of the range ofP.

The operatorU must be weakened to a partial isometry, rather than unitary, because of the following issues. IfA is theone-sided shift onl²(N), then |A| = {A^*A}^1/2 =I. So ifA =U |A|,U must beA, which is not unitary.

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

The operatorC can be defined byC(Bh) :=Ah for allh inH, extended by continuity to the closure ofRan(B), and by zero on the orthogonal complement to all ofH. The lemma then follows sinceA^*A ≤B^*B implies ker(B) ⊂ ker(A).

In particular. IfA^*A =B^*B, thenC is a partial isometry, which is unique if ker(B^*) ⊂ ker(C).In general, for any bounded operatorA,$${\displaystyle A^{\ast }A={\left(A^{\ast }A\right)}^{1/2}{\left(A^{\ast }A\right)}^{1/2},}$$where (A^*A)^1/2 is the unique positive square root ofA^*A given by the usualfunctional calculus. So by the lemma, we have$${\displaystyle A=U{\left(A^{\ast }A\right)}^{1/2}}$$for some partial isometryU, which is unique if ker(A^*) ⊂ ker(U). TakeP to be (A^*A)^1/2 and one obtains the polar decompositionA =UP. Notice that an analogous argument can be used to showA = P'U', whereP' is positive andU' a partial isometry.

WhenH 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 ofsingular value decomposition.

By property of thecontinuous functional calculus, |A| is in theC*-algebra generated byA. A similar but weaker statement holds for the partial isometry:U is in thevon Neumann algebra generated byA. IfA is invertible, the polar partU will be in theC*-algebra as well.

IfA is a closed, densely definedunbounded 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 asA, andU 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), andA^*Ah =B^*Bh for allh ∈ dom(A^*A), then there exists a partial isometryU such thatA =UB.U is unique if ran(B)^⊥ ⊂ ker(U). The operatorA being closed and densely defined ensures that the operatorA^*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 operatorA isaffiliated to a von Neumann algebraM, andA =UP is its polar decomposition, thenU is inM and so is the spectral projection ofP, 1_B(P), for any Borel setB in[0, ∞).

The polar decomposition ofquaternions${\displaystyle \mathbb{H}}$ withorthonormal basis quaternions${\displaystyle 1,\hat{ı},\hat{ȷ},\hat{k}}$ depends on the unit 2-dimensional sphere${\displaystyle \hat{r}\in \{x\hat{ı}+y\hat{ȷ}+z\hat{k}\in \mathbb{H}\setminus \mathbb{R}:x^2+y^2+z^2=1\}}$ ofsquare roots of minus one, known asright versors. Given any${\displaystyle \hat{r}}$ on this sphere and an angle−π <a ≤π, theversor${\displaystyle e^{a\hat{r}}=\cos a+\hat{r}\sin a}$ is on the unit3-sphere of${\displaystyle \mathbb{H}.}$ Fora = 0 anda =π, the versor is 1 or −1, regardless of whichr is selected. Thenormt of a quaternionq is theEuclidean distance from the origin toq. When a quaternion is not just a real number, then there is aunique polar decomposition:$${\displaystyle q=t\exp (a\hat{r}).}$$Herer,a,t are all uniquely determined such thatr is a right versor(r² = –1),a satisfies0 <a <π, andt > 0.

In theCartesian plane, alternative planarring decompositions arise as follows:

• Ifx ≠ 0,z =x(1 + ε(y/x)) is a polar decomposition of adual numberz =x +yε, whereε² = 0; i.e.,ε isnilpotent. In this polar decomposition, the unit circle has been replaced by the linex = 1, the polar angle by theslopey/x, and the radiusx is negative in the left half-plane.
• Ifx² ≠y², then theunit hyperbolax² −y² = 1, andits conjugatex² −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 thehyperbolic anglea and is written$${\displaystyle \cosh a+j\sinh a=\exp (aj)=e^{aj},}$$ wherej² = +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 byj reflects a point across the liney =x, the conjugate hyperbola has branches traced byje^aj or −je^aj. Therefore a point in one of the quadrants has a polar decomposition in one of the forms:$${\displaystyle r{e}^{aj},-r{e}^{aj},rj{e}^{aj},-rj{e}^{aj},r>0.}$$ The set{1, −1,j, −j} has products that make it isomorphic to theKlein four-group. Evidently polar decomposition in this case involves an element from that group.

Polar decomposition of an element of thealgebra M(2, R) of 2 × 2 real matrices uses these alternative planar decompositions since any planarsubalgebra is isomorphic to dual numbers, split-complex numbers, or ordinary complex numbers.

To compute an approximation of the polar decompositionA =UP, usually the unitary factorU is approximated.^[8][9] The iteration is based onHeron's method for the square root of1 and computes, starting from${\displaystyle U_0=A}$, the sequence$${\displaystyle U_{k+1}=\frac{1}{2}\left(U_k+{\left(U_k^{\ast }\right)}^{-1}\right),k=0,1,2,\dots }$$

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:

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

• Conway, J. B. (1990).A Course in Functional Analysis.Graduate Texts in Mathematics. New York: Springer.doi:10.1007/978-1-4757-4383-8.
• Douglas, R. G. (1966). "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space".Proc. Amer. Math. Soc.17:413–415.doi:10.1090/S0002-9939-1966-0203464-1.
• 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.

• Lie groups
• Operator theory
• Matrix theory
• Matrix decompositions

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