# 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.

• Apparently you were asked to show existence of a particular $QR$ factorization of $A$. Start by showing the $Q$ you have defined is orthogonal. Then come up with a way to define an upper triangular $R$ that makes $A=QR$. – hardmath Jan 16 at 13:04
• It's $Q^TQ=(A(L^T)^{-1})^T(A(L^T)^{-1})=L^{-1}A^TA(L^T)^{-1}=L^{-1}LL^T(L^T)^{-1}=I$, so $Q$ is orthogonal. Then $A=QL^T$ and $R=L^T$. Is that all to prove the existence of a QR factorization with $Q=A(L^T)^{-1}$ when $LL^T$ is the Cholesky decomposition of $A^TA$? – Tartulop Jan 16 at 16:51
• Yes, you are correct. If you know that nonsingular $A^T A = L L^T$ where $L$ is lower triangular, then you easily construct a $QR$ factorization just as your Comment shows. I suggest you write it up with a bit more detail as an Answer to your own Question. – hardmath Jan 16 at 17:28

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.