Polar decomposition

A real square ${\displaystyle m\times m}$ matrix ${\displaystyle A}$ can be interepreted as the linear 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 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 ${\displaystyle A}$ by a scale factor ${\displaystyle \sigma _{i}}$ (the action of ${\displaystyle P}$), followed by a single rotation or reflection 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.

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

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

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 ${\displaystyle A}$ is normal if and only if ${\displaystyle P'=P}$. Then ${\displaystyle U\Sigma =\Sigma U}$, and it is possible to diagonalise ${\displaystyle U}$ with a unitary similarity matrix ${\displaystyle S}$ that commutes with ${\displaystyle \Sigma }$, giving ${\displaystyle SUS^{*}=\Phi ^{-1}}$, where ${\displaystyle \Phi }$ is a diagonal unitary matrix of phases ${\displaystyle e^{i\varphi }}$. Putting ${\displaystyle Q=VS^{*}}$, one can then re-write the polar decomposition as

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

so ${\displaystyle A}$ 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 ${\displaystyle Q}$.

The polar decomposition of a square invertible real matrix ${\displaystyle A}$ is of the form

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

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

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

For any ${\displaystyle A}$, 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 such that ${\displaystyle A^{*}A=VDV^{*}}$, with ${\displaystyle D}$ diagonal and positive semi-definite.

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 ${\displaystyle \Lambda }$. 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 with containing the phases of the elements of ${\displaystyle \Lambda }$, that is, ${\displaystyle (\Phi _{\Lambda })_{ii}\equiv \Lambda _{ii}/|\Lambda _{ii}|}$ or ${\displaystyle (\Phi _{\Lambda })_{ii}}$ an arbitrary complex number with unit magnitude 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 with eigenvalues equal to the phases and absolute values of the eigenvalues of ${\displaystyle A}$, respectively.

From the singular-value decomposition, it can be shown that a ${\displaystyle A}$ 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.

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 ${\displaystyle A}$:

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