Existence of a QR factorization Let $A \in \mathbb{R^{n \times n}}$ be an nonsingular matrix and $LL^T$ the Cholesky decomposition of $A^TA$.
How to show that it exists a QR factorization with $Q=A(L^T)^{-1}$?
I tried this:
$A^TA=R^TQ^TQR=L^TL \Leftrightarrow (R^T)^{-1}A^TA=Q^TQR=(R^T)^{-1}L^TL$
Here I don't see how to continue. I tried to rearrange this equality, but I don't get the result.
 A: Let's unpack the solution that the OP arrived at in Comments, for the sake of completeness.  We assume $A$ is a nonsingular $n\times n$ (square) real matrix.
Suppose that $A^TA$ has Cholesky decomposition $LL^T$ where $L$ is lower triangular.  We will show that $A = QR$ is a factorization with $Q$ orthogonal and $R$ upper triangular when $Q=A(L^T)^{-1}$ and $R = L^T$.
First note that by our assumption $A^TA$ is symmetric positive definite and thus has the required Cholesky decomposition, and that $L$ is also nonsingular. The definition of $Q$ therefore makes sense, and the calculation that it is orthogonal is straightforward:
$$ \begin{align*} QQ^T &= A(L^T)^{-1} L^{-1} A^T \\
  &= A(LL^T)^{-1} A^T \\
  &= A(A^TA)^{-1} A^T \\
  &= AA^{-1}(A^T)^{-1} A^T \\
  &= I \end{align*} $$
Here we've used familiar facts that both inverse and transpose of a matrix product reverse the order of that product, and that inverse commutes with transpose.
Since $R = L^T$ is obviously upper triangular, it remains only to verify the factorization of $A$ that results:
$$ QR = A(L^T)^{-1} L^T = A $$
In this last calculation only the associativity of matrix multiplication is needed.
