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As stated, I want to show that if $A=UP$, where $P$ is the square root of $A^*\!\!A$ (with $*$ denoting adjoint) and $U$ is a unitary matrix, then $U$ is unique if and only if $A$ is invertible.

I am able to prove one direction : $A$ invertible implies $U$ is unique.

How do I prove the other direction?

Mohit Kumar
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  • What do you mean by "$P$ is the square root of $A$"? In polar decomposition, $P$ is almost always not a square root of $A$. Not to mention that square roots of a matrix is usually not unique. – user1551 Sep 21 '20 at 16:44
  • Incidentally, this was another confusion that I had with the polar value decomposition. I now understand that $P$ is $(A^*A)^{\frac{1}{2}}$. Have updated the question – Mohit Kumar Sep 22 '20 at 08:24

1 Answers1

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Suppose $\det\big(A\big)=0$. Then $\sigma_n =0$. It suffices to show that $\sigma_n=0$ implies two distinct possibilities for the unitary matrix in polar decomposition.

$A=UP = U\big(Q\Sigma Q^*\big)$

$D:=\begin{bmatrix} I_{n-1} & \mathbf 0 \\ \mathbf 0&-1 \end{bmatrix}$

consider unitary $S:=QDQ^*$
1.) $AS=\big(UQ\Sigma Q^*\big)S = UQ\big(\Sigma D\big) Q^*=UQ\big(\Sigma \big) Q^* = UP = A$
2.) $AS = UPS= USP = \big(US\big)P = U'P$
and for avoidance of doubt, $U'\neq U$ because $U$ is invertible

user8675309
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  • Thanks a lot! This may be slightly off topic, but what was your thought process to arrive at this answer? The argument doesn't seem something that I would naturally think of. – Mohit Kumar Sep 22 '20 at 13:59
  • (1) that $\sigma_n = 0$ is the deciding item for $A$ being singular, (2) consider the case where $P$ is diagonal, how would you show it then, (3) do a unitary similarity transform to cover the 'more general' case. – user8675309 Sep 22 '20 at 17:41