As it known, according to the Polar Decomposition, square matrix can be expressed in the next form $$M=QR$$ ($Q$ - orthogonal matrix $R$ - positive-semidefinite Hermitian matrix)

I need to find this $Q$ factor for the case of $3\times 3$ matrix. For this purpose I use next well known iterative formula

$$ Q_{i+1} = \frac{1}{2}\left[ Q_i+(Q_i^{-1})^T \right] $$

where the $\det Q_0\neq0 $ However, on practice it works a little bit slowly (it takes, more than 15 iteration before finds right answer). Is there any other, faster algorithms exist to perform Polar Decomposition? I have found exact formula for finding $Q$ factor for the case of $2\times 2$ matrices.

$$Q = M + \mathrm{sign}(\det M)\begin{pmatrix} d & -c\\ -b & a\\ \end{pmatrix}$$ where the initial matrix $$M=\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$$ Does such formula exist for $3\times 3$ matrix?

  • $\begingroup$ Maybe this question is a bit obvious, but just to be thorough: have you tried the LAPACK QR routine? $\endgroup$ Commented Nov 1, 2013 at 22:09
  • $\begingroup$ It's not the QR decomposition, the original poster just used the same letters. In the polar decomposition, the $R$ is positive semidefinite and not triangular. $\endgroup$
    – Nick Alger
    Commented Nov 5, 2013 at 1:10
  • 3
    $\begingroup$ A faster iteration than the one you are using is described in: Yuji Nakatsukasa and Nicholas J. Higham, Backward stability of iterations for computing the polar decomposition, SIAM Journal on Matrix Analysis and Applications, Vol. 33, No. 2, pp. 460-479, 2012. However, I am not sure if this is the best way if you are only working with $3\times 3$ matrices; if you find a (stable) closed form as suggested below it is probably much faster. $\endgroup$ Commented Nov 6, 2013 at 9:54

3 Answers 3


You can reduce the problem to computing the singular value decomposition, for which there exist many fast methods and codes. For fast 3x3 SVD, I found this paper.

To reduce the polar decomposition to the SVD, suppose the polar decomposition is written in the following form, $$M = U P,$$

with orthogonal $U$ and positive semidefinite $P$ (Ie., $Q \rightarrow U$, $R \rightarrow P$ from your notation). Further, denote the SVD of $P$ by $$P = V \Sigma V^T.$$

Substituting the second equation into the first yields, $$M = U V \Sigma V^T.$$

In other words, if you compute the SVD of $M$, $$M = W \Sigma V^T,$$

then $P$ is given by the above formula, and $$U = W V^T.$$

Incidentally, when you want to prove the existence of SVD-like decompositions for infinite dimensional operators, one basic strategy is to start with the polar decomposition which is easier to prove, and then do these steps backwards.

  • $\begingroup$ I think you need to use the conjugate transpose instead of the regular transpose. $\endgroup$
    – Azmisov
    Commented May 14, 2014 at 22:39
  • 1
    $\begingroup$ Sure, if your input matrix is complex-valued then you need the conjugate transpose. However, if it is real valued then $V$ is real-valued so it doesn't matter. (unlike the eigenvalue decomposition where real matrices could have complex eigenvectors) $\endgroup$
    – Nick Alger
    Commented May 14, 2014 at 23:32

Section 2.5 in Continuum Mechanics by A.J.M. Spencer is devoted to the 3x3 polar decomposition.


A more recent publication has come out with a new method for solving the 3x3 polar decomposition.

An algorithm to compute the polar decomposition of a 3x3 matrix

(I'm really surprised OP was needing 15 iterations for the iterative method!)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.