In section 2 of Pawe{\l} Zieli\'{n}ski, Krystyna Zi\c{e}tak: {The Polar Decomposition -- Properties, Applications and Algorithms}, Annals of the Polish Mathematical Society, vol.38, \bdate{1995}. %URL: \url{http://citeseer.ist.psu.edu/zielinski95polar.html}.} URL: \url{http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.50.3541}.} % http://www.im.pwr.wroc.pl/~zietak/papers/selected.html \jjfile{zielinski95polar.ps.gz} it is said (A=H*U, H Hermititian, U unitary) that "...the Hermititian factor H is always unique. ...the unitary factor U is unique if A has full rank" Higham's paper in my list touching sign and polar decomposition are Nicholas J.\ Higham: {The Matrix Sign Decomposition and its Relation to the Polar Decomposition}, Linear Algebra and its Applications, 212-213, pp.3-20, \bdate{1994}. %% sic, cf. http://eprints.ma.man.ac.uk/361/ URL: \url{http://www.maths.manchester.ac.uk/~higham/papers/matrix-functions.php}.} \jjfile{higham94matrix.ps.gz} {matrixsqrt}{Nicholas J.\ Higham: {Stable Iterations for the Matrix Square Root}, Numerical Algorithms, vol.15, no.2, pp.227-242, \bdate{1997}. URL: \url{http://www.maths.manchester.ac.uk/~higham/papers/matrix-functions.php}.} \jjfile{higham-stable-matrix-sqrt.ps.gz} * Fred lunnon <fred.lunnon@gmail.com> [Sep 09. 2010 09:47]:
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
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