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.

  • $\begingroup$ 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 $A=QR$. $\endgroup$ – hardmath Jan 16 at 13:04
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    $\begingroup$ 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 $A^TA$? $\endgroup$ – Tartulop Jan 16 at 16:51
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    $\begingroup$ 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. $\endgroup$ – 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.


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