Cholesky decompostion: upper triangular or lower triangular? Can I find and use $U$ such that
$$A = U U^{T}$$
where $U$ is an upper triangular matrix, to find a solution instead of finding $L$ such that $ A = L L^{T}$ (where $L$ is a lower triangular matrix) to solve $Ax=b$ using Cholesky factorization? If not, what is the correct way of using Cholesky factorization?
 A: You can use $A=UU^T$ to solve linear equations. However it is not what people call Cholesky decomposition. Also the algorithm is less intuitive. 
Note: Both decompositions are equivalent. Let $P$ the permutation which reverses the order, which is symmetric.  Then, $\tilde A = PAP$ is symmetric and positive definite iff $A$ is. In particular we have the cholesky decomposition
$$ \tilde A = PAP = (PUP)(PUP)^T. $$
A: We can modify the original Cholesky algorithm to obtain an upper-triangular version.
According to https://en.wikipedia.org/wiki/Cholesky_decomposition#The_Cholesky_algorithm, the original method starts at the top left and goes down to the bottom right by using this recursion step $A_i = L_i A_{i+1} L_i^T$, where $L_i$ is a lower-triangular matrix. The $A_i$ in the original version is created in a top-down fashion.
The upper-triangular version of the algorithm starts at the bottom right and goes up to the top left using a similar recursion $A_i = U_i A_{i+1} U_i^T$, where $U_i$ is an upper-triangular matrix. The $A_i$ matrix is created in a bottom-up fashion as
$
A_{i}=\begin{bmatrix}
B & b & 0 \\ b^T & a & 0 \\ 0 & 0 & I_{i-1} 
\end{bmatrix}.$
The $A_{i+1}$ matrix is defined as
$
A_{i+1}=\begin{bmatrix}
B-\frac{1}{a}bb^T & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & I_{i-1} 
\end{bmatrix}.$
The $U_i$ matrix is defined as
$
U_{i}=\begin{bmatrix}
I_{n-i} & \frac{1}{\sqrt{a}}b & 0 \\ 0 & \sqrt{a} & 0 \\ 0 & 0 & I_{i-1} 
\end{bmatrix}.$
We can check that this expression $A_i = U_i A_{i+1} U_i^T$ holds, which gives an upper-triangular version of the Cholesky algorithm ($A=UU^T$, where $U=U_1U_2\dots U_n$ is an upper-triangular matrix).
