Is Polar Decomposition Unique?

Question

For an arbitrary complex matrix $A$, it is known that $A$ has either a decomposition $P U$ or $V Q$, where $P = (A A^{*})^{1 / 2}$, $Q = (A^{*} A)^{1 / 2}$, and $U, V$ are isometry. Question: Is it possible for a matrix $A$ to have a decomposition $R W$ or $W R$ (where $R$ and $W$ are positive semidefinite and isometry factors, respectively) such that $R \neq P, Q$?

Answer 1

Suppose that $A=VQ=UP$, where $P,Q$ are positive and $U,V$ are isometries. Then $$ A^*A=PU^*UP=P^2. $$ So $P=(A^*A)^{1/2}$. The analog computation for $Q$ then shows that $Q=P$.

There is no uniqueness for the isometry though, without further requirements. For instance let $$ A=\begin{bmatrix}0&1\\0&0\end{bmatrix}. $$ Then $$ A=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix} $$ gives you two decompositions with different isometries.

It is common, to get uniqueness, to require instead that $U$ be a partial isometry with initial space the range of $A^*$ and final range the range of $A$. This can be stated as $U^*U=[A^*]$, $UU^*=[A]$, where the square brackets denote the range projection.