Let A be an $m\times n$ matrix of rank $n$. Let $$A = Q \begin{bmatrix}R\\ 0\end{bmatrix}$$ be the QR factorization of A, with $Q$ orthogonal and $R$ an $n\times n$ upper triangular matrix. Let $A^TA=LL^T$ be the Cholesky factorization of $A^TA$.
(a) Show that $R^TR=LL^T$.
(b) Can one conclude that $R=L^T$, if not, give a counterexample.
I have proved the first one, but I cannot show the (b). Is the Cholesky factorization unique?
Slightly longer but complete proof: I'm not going to show that the Cholesky factorisation exists and take that as done. To prove the uniqueness, we can proceed by induction. The statement is trivial for $n=1$, so assume that it is true for SPD matrices of the dimension $n-1$. Let $L$ be the Cholesky factor of $A$ with positive diagonal entries and assume that there is another such Cholesky factor $\tilde{L}$. Partition all these matrices as $$ A=\begin{bmatrix} A_{11} & a_{12} \\ a_{21}^T & a_{22} \end{bmatrix}, \quad L=\begin{bmatrix} L_{11} & 0 \\ l_{21}^T & l_{22} \end{bmatrix}, \quad \tilde{L}=\begin{bmatrix} \tilde{L}_{11} & 0 \\ \tilde{l}_{21}^T & \tilde{l}_{22} \end{bmatrix}, $$ where the blocks $A_{11}$, $L_{11}$, and $\tilde{L}_{11}$ are $(n-1)\times(n-1)$. Since $A=L_{11}L_{11}^T=\tilde{L}_{11}\tilde{L}_{11}^T$ are two Cholesky factorisations of the $(n-1)\times(n-1)$ SPD matrix $A_{11}$ (with positive diagonal), we have $\color{red}{\tilde{L}_{11}=L_{11}}$ from the induction assumption. Hence we have that $$ \tilde{L}=\begin{bmatrix} L_{11} & 0 \\ \tilde{l}_{21}^T & \tilde{l}_{22} \end{bmatrix} $$ and it remains to show that if $l_{22}$ and $\tilde{l}_{22}$ are positive, then $\tilde{l}_{21}=l_{21}$ and $\tilde{l}_{22}=l_{22}$. Since $LL^T=\tilde{L}\tilde{L}^T$, we have that $\hat{L}^T\hat{L}=I$ with $\hat{L}=\tilde{L}^{-1}L$. We can compute the matrix $\hat{L}$ "explicitly" in the block form as follows: $$ \begin{split} \hat{L}&=\tilde{L}^{-1}L \\&= \begin{bmatrix} L_{11} & 0 \\ \tilde{l}_{21}^T & \tilde{l}_{22} \end{bmatrix}^{-1} \begin{bmatrix} L_{11} & 0 \\ l_{21}^T & l_{22} \end{bmatrix} \\&= \begin{bmatrix} L_{11}^{-1} & 0 \\ -\tilde{l}_{22}^{-1}\tilde{l}_{21}^TL_{11}^{-1} & \tilde{l}_{22}^{-1} \end{bmatrix} \begin{bmatrix} L_{11} & 0 \\ l_{21}^T & l_{22} \end{bmatrix} \\&= \begin{bmatrix} I & 0 \\ \tilde{l}_{22}^{-1}(l_{21}-\tilde{l}_{21})^T & \tilde{l}_{22}^{-1}l_{22} \end{bmatrix}. \end{split} $$ Therefore, $$ \begin{split} I&=\hat{L}\hat{L}^T \\&= \begin{bmatrix} I & 0 \\ \tilde{l}_{22}^{-1}(l_{21}-\tilde{l}_{21})^T & \tilde{l}_{22}^{-1}l_{22} \end{bmatrix} \begin{bmatrix} I & \tilde{l}_{22}^{-1}(l_{21}-\tilde{l}_{21}) \\ 0 & \tilde{l}_{22}^{-1}l_{22} \end{bmatrix} \\&= \begin{bmatrix} I & \tilde{l}_{22}^{-1}(l_{21}-\tilde{l}_{21}) \\ \tilde{l}_{22}^{-1}(l_{21}-\tilde{l}_{21})^T & \tilde{l}_{22}^{-2}(l_{21}-\tilde{l}_{21})^T(l_{21}-\tilde{l}_{21})+\tilde{l}_{22}^{-2}l_{22}^2 \end{bmatrix}. \end{split} $$ Since the (2,1) and (1,2) block entries must be zero (the matrix on the right-hand side must be equal to the identity) and $\tilde{l}_{22}\neq 0$, we have $\color{red}{\tilde{l}_{21}=l_{21}}$. So what remains from the bottom-right entry of $\hat{L}\hat{L}^T$ is $\tilde{l}_{22}^{-2}l_{22}^2$ and this must be equal to 1. Because $\tilde{l}_{22}>0$ and $l_{22}>0$, we get $$ \tilde{l}_{22}^{-2}l_{22}^2=1\Leftrightarrow \tilde{l}_{22}^2=l_{22}^2\Leftrightarrow \color{red}{\tilde{l}_{22}=l_{22}}. $$ Therefore, $\color{blue}{\tilde{L}=L}$. Q.E.D.
To illustrate that the positivity of the diagonal of $L$ is a necessary condition for the uniqueness, consider the following example: $$ A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix}, \quad L=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}, \quad \tilde{L}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & -1 & 0 \\ 1 & -1 & 1 \end{bmatrix}. $$ Both $L$ and $\tilde{L}$ are Cholesky factors of $A$. However, $L$ is the only Cholesky factor of $A$ with positive diagonal entries.