# Polar decomposition for quaternionic matrices?

A non-zero complex number can be uniquely written in polar form as $re^{i\theta}$. There is an analogous result for complex matrices: any invertible complex matrix can be uniquely written as $UP$, where $U$ is a unitary matrix and $P$ is a positive definite Hermitian matrix (see e.g. the description on Wikipedia)

Suppose we consider $n\times n$ invertible matrices over the quaternions. Is there an analogous polar decomposition?

Or to ask the question a little more colorfully, can we complete the following list?

• roots of unity: positive reals
• unitary matrices: positive definite matrices
• compact symplectic matrices: ???

PS: Note that there exists a polar decomposition for quaternions; I'm interested in the result for matrices of quaternions.

A key wordphrase here is "Cartan decomposition". Since $G=SL_n(\mathbb H)$ is a semisimple group, there is a diffeomorphism $$K\times \mathfrak p\rightarrow G$$ taking $(k,X)$ to $k\cdot {\rm exp}(X)$, where $K$ is a maximal compact subgroup of $G$ (i.e. the compact symplectic group) and $\mathfrak p$ is the vector subspace of ${\mathfrak sl}_n(\mathbb H)$ fixed by the Cartan involution $X\rightarrow \bar X^t$ (the quaternionic version of Hermitian).

• Little correction: you have to take $GL$ and $\mathfrak{gl}$, since determinants do not exist for matrices over skew-fields. – Johannes Ebert Feb 16 '12 at 10:09
• Johannes, there is the extended notion of "reduced norm" (and "reduced trace") for simple algebras, which allows us to define $SL$ and $\mathfrak sl$, but I should have changed it to $GL$, since that is what the OP asks for (though I don't think the difference is worth editing in). – B R Feb 16 '12 at 14:20
• @BR Actually there is considerable confusion in the literature about what $SL_n(\mathbb{H})$ is. The definition varies from one source to the next, depending on how the extended determinant is defined, and depending on whether one includes additional conditions to interpret it as an isometry group. – j0equ1nn Mar 1 '17 at 5:17

For all matrices, even non-invertible matrices, one gets a polar decomposition $X=UP$ with $U$ unitary and $P$ positive semidefinite. Of course $U$ is not unique unless $X$ is invertible. I found this result in "Quaternions and matrices of quaternions" by F. Zhang, Linear Algebra Appl., 251 (1997), pp. 21–57.

There also the Jordan canonical form, Schur factorization and the spectral theorem. I have a survey on this that needs some polishing: http://arxiv.org/abs/1107.0500 "Factorization of Matrices of Quaternions."

• I did a little light editing so that this actually answers the question. – Terry Loring Feb 17 '12 at 19:00

The good news is that this is addressed in:

Zhuang, Wa Jin(PRC-ZNU) Polar decomposition and GL partial ordering for quaternion rectangular matrices. (Chinese. English, Chinese summary) Adv. Math. (China) 34 (2005), no. 2, 187–193.

The bad news is that I can't seem to find the paper (but the math review is reasonably informative).

I guess just as for normal SVD, you can recover the polar decomposition for quaternion matrices from their corresponding SVD, for which two immediate papers that I found were: