I am learning polar decomposition.
In the examples I have seen (for example http://buzzard.ups.edu/courses/2014spring/420projects/math420-UPS-spring-2014-buffington-polar-decomposition.pdf concrete calculations starting page 4), a first step is to calculate $A^*A$. Then this is eigendecomposed into $U^{-1} D U$ with $D$ diagonal and $U$ unitary (my choice of inverse $U$ term is for later). Because $D$ is diagonal it's easy to calculate the square root, then it can be substituted: $A^*A=U^{-1}\sqrt{D}\sqrt{D}U$. If we were given $A$ as an matrix in a basis then all the terms so far have explicit numerical values from the eigendecomposition algorithm. Now at this point the reference procedures go on and do a few more steps.
However, I see that with $U$ unitary $U^*=U^{-1}$ and with $D$ diagonal $\sqrt{D}$ is also diagonal and $\sqrt{D}^*=\sqrt{D}$. So $A^*A=U^*\sqrt{D}^*\sqrt{D}U=(\sqrt{D}U)^*(\sqrt{D}U)$. So simply $A = \sqrt{D}U$ and the decomposition is solved. Things make me doubtful of this analysis:
- $\sqrt{D} U$ is the polar decomposition directly. If it was possible to just read off the results at this stage, why would the published algorithms do more steps?
- I thought maybe my reasoning is a special case that does not work in all cases. However, in the examples I have seen, which seem quite "clean" (all real numbers, not apparently singular, etc.) it seems like the final results of the following calculations are different from these matrices. So, it's not just my analysis is a special case, it is wrong.
- $\sqrt{D}$ is positive definite diagonal. Maybe polar decomposition can break down into Unitary and Hermitian (actually I think sometimes not), but I don't think it can break down into Unitary and Diagonal, that is too strong.
I think maybe $D$ if it is diagonal has complex eigenvalues and hence it is not Hermitian, which causes the problem. But everwhere I have read about this procedure it seems to be positive semidefinite.
Can you please let me know where my reasoning is wrong?
For matrices $A,B$ of the same shape, the fact that $A^*A = B^*B$ does not imply that $A = B$. Thus, the fact that $A^*A = (\sqrt{D}U)^*(\sqrt{D}U)$ does not imply that $A = \sqrt{D}U$.
To see that in general this will not be the case, note that the matrix $B = \sqrt{D} U$ will always have the property that $BB^*$ is diagonal, which is to say that the rows of $B$ are orthogonal. However, it is certainly possible that $A$ does not have orthogonal rows.