memo

Polar Decomposition

Polar decomposition of $A$ is a decompositionof the form

\[A = UP\]

where $U \in \mathbb{C}^{n \times n }$ unitary matrix and $P$ is a positive-semidifinite hermitian matrix.

Then following statements are equivalent:

(a) $\Rightarrow$ (b) Suppose $A = W \Sigma V^{*}$. Let

\[\begin{eqnarray} P & := & V \Sigma V^{*} \nonumber \\ U & := & WV^{*} . \nonumber \end{eqnarray}\]

$U$ is unitary matrix since

\[\begin{eqnarray} U^{*}U & = & VW^{*}WV^{*} \nonumber \\ & = & I . \nonumber \end{eqnarray}\]

Diagonal matrix $\Sigma$ is positve semidefinite since

\[\begin{eqnarray} x^{\mathrm{T}}\Sigma x = \sum_{i=1}^{r} d_{i}(x_{i})^{2} \ge 0 \end{eqnarray}\]

$P$ is positve definite since

\[\begin{eqnarray} x^{*}V\Sigma V^{*}x & = & (V^{*}x)^{*}\Sigma (V^{*}x) \nonumber \end{eqnarray}\]

(a) $\Leftarrow$ (b)

$P$ is symmetric positive semidefinite so that by eigenvalue decomposition weh have

\[P = VDV^{*}\]

where $V$ unitary matrix and $D = \mathrm{diag}(d_{1}, \ldots, d_{r}, 0, \ldots, 0)$ and $d_{i} > 0$. Then we define

\[\begin{eqnarray} A & = & UP \nonumber \\ & = & UVDV^{*} \nonumber \\ & = & WDV \nonumber \end{eqnarray}\]

where $W := UV$. $W$ is unitary matrix since $U$ and $V$ are both unitary.

$\Box$

Then polar decomposition is unique and $P$ is positive definite.

Suppose that there exists another decomposition $A = U_{1}P_{1}$ where $U_{1}$ is unitary and $P_{1}$ is hermitian positive semidefinite matrix.

\[A^{\mathrm{T}}A = P_{1}^{\mathrm{T}}P\]

TBD

$\Box$