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Is there a way to simplify block Cholesky decomposition If you already have decomposed the sub matrices along the leading diagonal?

Lets say we have a block matrix $ M =\left( \begin{array}{ccc} A & B\\\\ B^{*} & C \end{array} \right)$ where M is positive definite. (A, and C are also pos def)

There is a formula for carrying out block Cholesky decomposition. See http://en.wikipedia.org/wiki/Block_LU_decomposition. Summarising we have the following result.

The matrix $<math>\begin{matrix}M = LU\end{matrix}</math>$ can be decomposed in an algebraic manner into $<math>L = \begin{pmatrix} A^{\frac{1}{2}} & 0 \\\\ B^{\*} A^{-\frac{*}{2}} & Q^{\frac{1}{2}} \end{pmatrix}$

where

$\begin{matrix} Q = C - B^{*} A^{-1} B \end{matrix}$

$\*$ indicates transpose in this case

Now lets say we have already carried out the cholesky decomposition for A, and C. So we have already calculated $A^{1/2}$, and $C^{1/2}$ (It is therefore straightforward to calculate the inverses $A^{-1/2}$, and $C^{-1/2}$ using forward substitution).

Rewriting the Q in terms of these quantities we now have.

$Q = Q^{1/2}Q^{\*/2} = C^{1/2} C^{\*/2} - (B^{*} A^{-\*/2})(A^{-1/2} B)$

My question is this: Given this set up is it possible to algebraicly calculate $Q^{1/2}$ without having to apply cholesky decomposition to $Q$. Or in other words can I use $C^{1/2}$ to help me in the calculation of $Q^{1/2}$.

Thanks in advance for any replies.

Matt.