<s>[INST] Cholesky decomposition of the inverse of a matrix 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.
 [/INST] 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
$$X^{-1}=
\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 -\frac{4}{73} & 1 & 0 \\
 -\frac{20}{73} & \frac{2}{15} & 1 \\
\end{array}
\right),\left(
\begin{array}{ccc}
 \frac{73}{650} & 0 & 0 \\
 0 & \frac{15}{146} & 0 \\
 0 & 0 & \frac{1}{15} \\
\end{array}
\right),\left(
\begin{array}{ccc}
 1 & -\frac{4}{73} & -\frac{20}{73} \\
 0 & 1 & \frac{2}{15} \\
 0 & 0 & 1 \\
\end{array}
\right)
$$
notebook
</s>