[math-fun] Polar Decomposition
In numerous recent publications, it is claimed that the "polar decomposition" of a (real) matrix M = Q S as the product of orthonormal ("orthogonal") and symmetric factors is unique, at any rate provided M is nonsingular. See for example http://en.wikipedia.org/wiki/Polar_decomposition or the very readable (and available for free download) Ken Shoemake, Tom Duff "Matrix Animation and Polar Decomposition" As it stands, the claim is obviously false. A trivial example is M = [[0,1],[1,0]] which factorises as M I or as I M. There is no mention of uniqueness on page 6 of Gantmacher "Theory of Matrices" vol 2 though he does point out that the factors commute if M commutes with its transpose, as in the counterexample above. The claim of uniqueness may be possibly result from confusion connected with the Higham 1986 reference in Shoemake, where there is an algorithm giving Q such that Q - M has minimal 2-norm. Can anyone cast light on this? Is the decomposition unique under the minimality constraint? Are there less onerous constraints guaranteeing uniqueness (and minimality)? Have I simply misunderstood something, or is this another rhinoceros' pancreas? Fred Lunnon
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 On Wed, Sep 8, 2010 at 6:26 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
In numerous recent publications, it is claimed that the "polar decomposition" of a (real) matrix M = Q S as the product of orthonormal ("orthogonal") and symmetric factors is unique, at any rate provided M is nonsingular.
See for example http://en.wikipedia.org/wiki/Polar_decomposition or the very readable (and available for free download) Ken Shoemake, Tom Duff "Matrix Animation and Polar Decomposition"
As it stands, the claim is obviously false. A trivial example is M = [[0,1],[1,0]] which factorises as M I or as I M.
There is no mention of uniqueness on page 6 of Gantmacher "Theory of Matrices" vol 2 though he does point out that the factors commute if M commutes with its transpose, as in the counterexample above.
The claim of uniqueness may be possibly result from confusion connected with the Higham 1986 reference in Shoemake, where there is an algorithm giving Q such that Q - M has minimal 2-norm.
Can anyone cast light on this? Is the decomposition unique under the minimality constraint? Are there less onerous constraints guaranteeing uniqueness (and minimality)?
Have I simply misunderstood something, or is this another rhinoceros' pancreas?
Fred Lunnon
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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
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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The quotation [other mail] of Pawe{\l} Zieli\'{n}ski seems to be straight out of Higham/Schreiber: {Fast polar decomposition of an arbitrary matrix}, SIAM J. Sci. Stat. Comput., 11(4):648--655, July 1990. near bottom of the web page http://www.maths.manchester.ac.uk/~higham/papers/matrix-functions.php First page: "The decomposition always exists, H is the unique Hermitian positive semidefinite square root of A*A (i.e., H (A*A)^{I/2}), and U is unique if and only if A has full rank (these properties are proved in [section] 2)." [...]
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Fred lunnon -
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Victor Miller