I have the Cholesky decomposition of a matrix $M$. However, I need the Cholesky decomposition of the inverse of the matrix, $M^{-1}$. Is there a fast way to do this, without first computing $M^{-1}$? In other words, is there a relationship between the Cholesky decompositions of a matrix and of its inverse?
My matrix is a covariance matrix and, hence, positive-definite.
On
To add to previous answers, if we view $X$ as covariance matrix of data, the relationship between two decompositions reduces to relationship between coefficients of "right-to-left" autoregressive model and "left-to-right" one (details)
In general, switching order of "inverse" and "Cholesky" gives different results. IE
$$ \text{X=}\left( \begin{array}{ccc} 10 & 0 & 4 \\ 0 & 10 & -2 \\ 4 & -2 & 15 \\ \end{array} \right) $$
Using Cholesky and inverting factors, gives this decomposition
$$X^{-1}= \left( \begin{array}{ccc} 1 & 0 & -\frac{2}{5} \\ 0 & 1 & \frac{1}{5} \\ 0 & 0 & 1 \\ \end{array} \right)\left( \begin{array}{ccc} \frac{1}{10} & 0 & 0 \\ 0 & \frac{1}{10} & 0 \\ 0 & 0 & \frac{1}{13} \\ \end{array} \right)\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{2}{5} & \frac{1}{5} & 1 \\ \end{array} \right) $$
While Cholesky on the inverse directly gives
$$X^{-1}= \left( \begin{array}{ccc} 1 & 0 & 0 \\ -\frac{4}{73} & 1 & 0 \\ -\frac{20}{73} & \frac{2}{15} & 1 \\ \end{array} \right),\left( \begin{array}{ccc} \frac{73}{650} & 0 & 0 \\ 0 & \frac{15}{146} & 0 \\ 0 & 0 & \frac{1}{15} \\ \end{array} \right),\left( \begin{array}{ccc} 1 & -\frac{4}{73} & -\frac{20}{73} \\ 0 & 1 & \frac{2}{15} \\ 0 & 0 & 1 \\ \end{array} \right) $$
If $L^T L = R$ is the available Cholesky decomposition, then inverting both sides of the equation you get,
$$L^{-1}(L^{T})^{-1} = R^{-1} $$
And since transposition and inverse are interchangeable:
$$L^{-1}(L^{-1})^{T} = R^{-1} $$
So if you define $P = (L^{-1})^T$ this is your desired answer. In other words,
$$P^{T}P=R^{-1}$$