Cholesky factors of covariance and precision matrix Let $\Sigma$ be a covariance matrix (symmetric positive-definite), and $\Omega = \Sigma^{-1}$ the corresponding precision matrix, which is also SPD (the quotients of positive eigenvalues are positive).
Let $T$ and $U$ be the upper-triangular Cholesky factors respectively of $\Sigma$ and $\Omega$, such that:
$$
\Sigma = T^t T \qquad\text{and}\qquad \Omega = U^t U
$$
writing down the inverse of $\Sigma$ explicitly, we also have:
$$
\Sigma^{-1} = T^{-1} T^{-t} = \Omega = U^t U
$$
and similarly the other way around, writing down the inverse of $\Omega$.
In this equality, we have a product of upper-times-lower triangular matrices on the left $T^{-1} T^{-t}$, and a product of lower-times-upper on the right $U^t U$. I get confused with such equations, and am not sure what to do next; considering products like $\Sigma\Omega$ do not lead anywhere (that I can see). Can we say anything about how $T$ and $U$ are related?
$\Sigma^{-1} = T^{-1} (T^{*})^{-1} = R^*R\implies I= TR^*RT^*=\big(RT^*\big)^*\big(RT^*\big)\implies Q^* = RT^*$
for some unitary $Q$
Multiply each side on the left by $Q$ and on the right by $(T^*)^{-1}$ to get
$(T^*)^{-1} = QR$
That is, they are related by the unique QR factorization of $(T^*)^{-1}$ such that $R$ has positive diagonals (uniqueness follows since the Cholesky factorization is unique for PD matrices).
(Could not edit the answer of @user8675309, so here is a cleaner derivation.)
The main equality relating $T$ and $U$ is:
$$ T^{-1} T^{-t} = U^t U $$
Multiplying by $T$ on the left, and $T^t$ on the right, the left-hand side becomes the identity:
$$ I = T U^t\ U T^t = (U T^t)^t\ UT^t $$
so the matrix $Q=UT^t$ is orthogonal. We can then write:
$$ T^{-t} = Q^t\ Q\ T^{-t} = Q^t\ UT^t\ T^{-t} = Q^t\ U $$
where we &quot;recognize&quot; the QR decomposition of $T^{-t}$. Uniqueness follows since the Cholesky factorization is unique for PD matrices.