# The Unapologetic Mathematician

## The Uniqueness of Polar Decomposition

I asserted in a comment that the polar decomposition of a transformation has certain uniqueness properties. First, we can always uniquely recover the positive-semidefinite part, and then if that’s actually positive-definite we can recover the unitary part. And it turns out our running analogy with complex numbers is the perfect way to get at these facts.

Recall that if we take a complex number in polar form $z=re^{i\theta}$ we can recover its length as

$\displaystyle\lvert z\rvert=\sqrt{\bar{z}z}=\sqrt{re^{-i\theta}re^{i\theta}}=\sqrt{r^2}=r$

Similarly, using the right polar decomposition we can write

$\displaystyle\sqrt{T^*T}=\sqrt{\left(UP\right)^*\left(UP\right)}=\sqrt{PUU^*P}=\sqrt{PP}=P$

where the radical denotes the unique positive-semidefinite square root any positive-semidefinite transformation has.

Now if $T$ is nonsingular then so is $P$, since we can write $P=U^*T$ and $U^*$ is definitely nonsingular. So in this case we can write $U=TP^{-1}$ and $U$ is uniquely determined. On the other hand, if $T$ is singular then we can’t invert $P$. We then have some freedom in our choice of $U$, corresponding to this nontrivial cokernel of $P$.

We could also use the left polar decomposition to write

$\displaystyle\sqrt{TT^*}=\sqrt{\left(P'U\right)\left(P'U\right)^*}=\sqrt{P'UU^*P'}=\sqrt{P'P'}=P'$

So the positive-semidefinite part is again uniquely determined, and again the unitary part will be uniquely determined so long as $T$ is nonsingular.

August 21, 2009