Cholesky factors of covariance and precision matrix

Question

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?

Answer 1

(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 "recognize" the QR decomposition of $T^{-t}$. Uniqueness follows since the Cholesky factorization is unique for PD matrices.

Answer 2

$\Sigma^{-1} = T^{-1} (T^{*})^{-1} = R^*R\implies I= TR^*RT^*=\big(RT^*\big)^*\big(RT^*\big)\implies Q^* = RT^*$
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).

Answer 3

Another, more direct answer, can be obtained by defining $V = T^{-1}$, which is upper-triangular, and writing down the actual product (notice the extents of each sum): $$ \forall\ n\geq i\geq j\geq 1,\quad \omega_{ij} = \sum_{k=i}^n v_{ik} v_{jk} = \sum_{k=1}^j u_{ki} u_{kj} $$

From this formula, we get an algorithm to compute $V$ as a function of $U$ with two for loops:

for i = n:-1:1
for j = i:-1:1

i.e. starting from the bottom-right and proceeding with backwards indexing row by row, and the relations: $$ \begin{aligned} v_{ii} &= \sqrt{ \sum_{k=1}^i u_{ki}^2 - \sum_{k=i+1}^n v_{ik}^2 } \\ \forall j<i,\quad v_{ji} &= \frac{ \sum_{k=1}^j u_{ki} u_{kj} - \sum_{k=i+1}^n v_{ik} v_{jk} }{v_{ii}} \end{aligned} $$