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I'm (reasonably) familiar with factoring a positive definite matrix $\mathbf{P} = \mathbf{L} \mathbf{L}^T = \mathbf{R}^T \mathbf{R}$, and is supported by MATLAB and Eigen.

However, I have also seen a factorization of the (same) \mathbf{P} = \mathbf{U} \mathbf{U}^T = \mathbf{L'}^T \mathbf{L'}

The following illustrates:

>> A = rand(3, 4)  A =      0.2785    0.9649    0.9572    0.1419     0.5469    0.1576    0.4854    0.4218     0.9575    0.9706    0.8003    0.9157  >> P = A * A.'  P =      1.9449    0.8288    2.0991     0.8288    0.7374    1.4513     2.0991    1.4513    3.3379  >> R = chol(P)  R =      1.3946    0.5943    1.5052          0    0.6198    0.8982          0         0    0.5153  % This function computes such that U * U.' = A * A.' % Part of: http://www.iau.dtu.dk/research/control/kalmtool2.html  >> U = triag(A)  U =     -0.7475    0.2571   -1.1489          0   -0.3262   -0.7944          0         0   -1.8270  >> P2 = R.' * R  P2 =      1.9449    0.8288    2.0991     0.8288    0.7374    1.4513     2.0991    1.4513    3.3379  >> P3 = U * U.'  P3 =      1.9449    0.8288    2.0991     0.8288    0.7374    1.4513     2.0991    1.4513    3.3379 

I haven't seen this particular factorization $\mathbf{P} = \mathbf{U} \mathbf{U}^T$ before. I have a couple of questions:

  • Is it still, by definition, Cholesky factoriation? If not, what is it called?
  • Is the simple means to compute this particular variant (e.g. a MATLAB command)
  • Is there a specific relationship between $\mathbf{U}$ and $\mathbf{R}$?

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You other item is not called Cholesky.

The Cholesky decomposition is unique: given a Hermitian, positive-definite matrix A, there is only one lower triangular matrix L with strictly positive diagonal entries such that A = LL*.

The quote is from http://en.wikipedia.org/wiki/Cholesky_decomposition

Cholesky is an example in the LU pattern, lower triangular on the left and upper triangular on the right, http://en.wikipedia.org/wiki/LU_decomposition

Wherever you found you code, it has switched the order. All you really know about your $U$ are things along the lines of U^{-1} R' is an orthogonal matrix, but as it is not upper or lower triangular you must work at it to find anything interesting. Also, of course, the absolute values of the determinants of $U$ and $R$ are equal.

I have not worked up a complete proof yet, but it appears that your $-U$ is likely to be unique as well, factor in reverse order of Cholesky and demand strictly positive entries on the diagonal, all assuming we are factoring a positive definite symmetric.

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    @Damien, simply from $U U^T = R^T R$ it follows that there is an orthogonal matrix $O$ with $U^{-1} R^T = O,$ or $R^T = UO.$ Fro the fact that both ordinary and "backwards" Cholesky are unique, there are presumably many relationships.2012-02-19