But thinking a little more (while studiously avoiding actually getting to grips with any actual proof!), I reckon now that ambiguities in the (nonsingular) SVD seem to actually cancel out, so that you wind up with the same polar decomp. whichever SVD you choose --- is this right? WFL On 9/9/10, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, Here might be the source of confusion. The matrix M has a singular value decomposition
M = U D V
where U, V are orthogonal and D is a diagonal matrix with the nonnegative values in the upper left hand and sorted in descending order. Such a D is unique, however, if diagonal values of D are equal, the U and V are not unique (since you can just interchange them and appropriately modify U and V). The relation between this and the polar decomposition is that a symmetric matrix S can be written in the form U D U^-1, which is almost unique (the same problem arises when two diagonal values of D are equal or opposite sign. So take the above singular value decomposition and write
M = (U D U^-1) (U V).
Note that the first factor is symmetric (and all symmetric factors arise this way) and the second factor orthogonal.
Victor