Editing Polar decomposition (section)

== 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]].

=== Derivation for normal matrices ===
If <math>A</math> is [[Normal matrix|normal]], then it is unitarily equivalent to a diagonal matrix: <math>A = V\Lambda V^*</math> for some unitary matrix <math>V</math> and some diagonal matrix <math>\Lambda ~.</math> This makes the derivation of its polar decomposition particularly straightforward, as we can then write
<math display="block">A = V\Phi_\Lambda|\Lambda|V^* = \underbrace{\left( V\Phi_\Lambda V^* \right)}_{\equiv U}\underbrace{\left(V|\Lambda|V^* \right)}_{\equiv P},</math>

where <math>|\Lambda|</math> is the matrix of absolute diagonal values, and <math>\Phi_\Lambda</math> is a diagonal matrix containing the ''phases'' of the elements of <math>\Lambda,</math> that is, <math>(\Phi_\Lambda)_{ii}\equiv \Lambda_{ii}/ |\Lambda_{ii}|</math> when <math>\Lambda_{ii} \neq 0,</math>, and <math>(\Phi_\Lambda)_{ii} = 0</math> when <math>\Lambda_{ii} = 0 ~.</math>

The polar decomposition is thus <math>A=UP,</math> with <math>U</math> and <math>P</math> diagonal in the eigenbasis of <math>A</math> and having eigenvalues equal to the phases and absolute values of those of <math>A,</math> respectively.

=== Derivation for invertible matrices ===
From the [[singular-value decomposition]], it can be shown that a matrix <math>A</math> is invertible if and only if <math>A^* A</math> (equivalently, <math>AA^*</math>) is. Moreover, this is true if and only if the eigenvalues of <math>A^* A</math> are all not zero.<ref>Note how this implies, by the positivity of <math>A^* A</math>, that the eigenvalues are all real and strictly positive.</ref>

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

In this expression, <math>V^*</math> is unitary because <math>V</math> is. To show that also <math>AVD^{-1/2}</math> is unitary, we can use the [[singular-value decomposition|SVD]] to write <math>A = WD^{1/2}V^*</math>, so that 
<math display="block">AV D^{-1/2} = WD^{1/2}V^* VD^{-1/2} = W,</math>
where again <math>W</math> is unitary by construction.

Yet another way to directly show the unitarity of <math>A\left(A^* A\right)^{-1/2}</math> is to note that, writing the [[singular-value decomposition|SVD]] of <math>A</math> in terms of rank-1 matrices as <math display="inline">A = \sum_k s_k v_k w_k^*</math>, where <math>s_k</math>are the singular values of <math>A</math>, we have 
<math display="block">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^*,</math> 
which directly implies the unitarity of <math>A\left(A^* A\right)^{-1/2}</math> 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 derivation ===
The SVD of a square matrix <math>A</math> reads <math>A = W D^{1/2} V^*</math>, with <math>W, V</math> unitary matrices, and <math>D</math> a diagonal, positive semi-definite matrix. By simply inserting an additional pair of <math>W</math>s or <math>V</math>s, we obtain the two forms of the polar decomposition of <math>A</math>:<math display="block">
  A = WD^{1/2}V^* =
  \underbrace{\left(W D^{1/2} W^*\right)}_P \underbrace{\left(W V^*\right)}_U =
  \underbrace{\left(W V^*\right)}_U \underbrace{\left(VD^{1/2} V^*\right)}_{P'}.
</math>More generally, if <math> A </math> is some rectangular <math>
  n\times m </math> matrix, its SVD can be written as <math> A=WD^{1/2}V^* </math> where now <math> W </math> and <math> V </math> are isometries with dimensions <math> n\times r </math> and <math>
 m\times r </math>, respectively, where <math> r\equiv\operatorname{rank}(A) </math>, and <math> D </math> is again a diagonal positive semi-definite square matrix with dimensions <math> r\times r </math>. We can now apply the same reasoning used in the above equation to write <math> A=PU=UP'</math>, but now <math> U\equiv WV^* </math> is not in general unitary. Nonetheless, <math> U </math> has the same support and range as <math> A </math>, and it satisfies <math> U^* U=VV^* </math> and <math> UU^*=WW^* 
</math>. This makes <math> U </math> into an isometry when its action is restricted onto the support of <math> A </math>, that is, it means that <math> U </math> is a [[partial isometry]].

As an explicit example of this more general case, consider the SVD of the following matrix:<math display="block">
  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}\sqrt2&0\\0&\sqrt8\end{pmatrix}}_{\sqrt D}
\underbrace{\begin{pmatrix}\frac1{\sqrt2} & \frac1{\sqrt2} \\ \frac1{\sqrt2} & -\frac1{\sqrt2}\end{pmatrix}}_{V^\dagger}. 
</math>We then have<math display="block">
  WV^\dagger = \frac1{\sqrt2}\begin{pmatrix}1&1 \\ 1&-1 \\ 0&0\end{pmatrix} 
</math>which is an isometry, but not unitary. On the other hand, if we consider the decomposition of<math display="block">
  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}, 
</math>we find<math display="block">
  WV^\dagger =\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix},  
</math>which is a partial isometry (but not an isometry).