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 $M = LU$ can be decomposed in an algebraic manner into

$L = \begin{pmatrix} A^{\frac{1}{2}} & 0 \\ B^{*} A^{-\frac{*}{2}} & Q^{\frac{1}{2}} \end{pmatrix}$

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.

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

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.

Let's 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 positive definite.)

There is a formula for carrying out block Cholesky decomposition. See Wikipedia: Block LU decomposition. Summarising we have the following result.

The matrix $M = LU$ can be decomposed in an algebraic manner into

$$L = \begin{pmatrix} A^{\frac{1}{2}} & 0 \\ B^{*} A^{-\frac{*}{2}} & Q^{\frac{1}{2}} \end{pmatrix}$$

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 $Q$ in terms of these quantities we now have

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

My question is this: Given this set up is it possible to algebraically 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 $Q^{1/2}$.

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.

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

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.

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 $M = LU$ can be decomposed in an algebraic manner into

$L = \begin{pmatrix} A^{\frac{1}{2}} & 0 \\ B^{*} A^{-\frac{*}{2}} & Q^{\frac{1}{2}} \end{pmatrix}$

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.

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

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.

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.

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.

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

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.