Suppose $M$ is an invertible matrix with a $QR$ decomposition. Because $M$ is invertible this $QR$ decomposition will be unique. I want to show it is unique via a least squares problem.
Say $Ax = b$ and as $A$ is invertible we will have an exact solution for the least squares problem. This (I think) should ultimately lead to showing $Q$ and $R$ are unique in the $QR$ decomposition of $A$. An exact solution means there exists $x$ such that $\Vert Ax - b \Vert_2 = 0$. Rewriting as
$$\Vert QRx - b \Vert_2 = 0$$
Ortogonal transformations are invariant under the 2-norm so multiplying by $Q^T$ gives
$$\Vert Rx - Q^Tb \Vert_2 = 0$$
Well that's as far as I got. I don't know if I'm even on the right track. Any ideas how to show the $QR$ decomposition is unique (via least squares) or is this actually not possible?
Matrix Computations Golub and Van Loan, 3e p. 230
Theorem 5.2.2
Suppose $\mathbf{A}\in\mathbb{R}^{m\times n}$ has full column rank. The thin $\mathbf{Q}\mathbf{R}$ factorization $$ \mathbf{A} = \mathbf{Q}_{1}\mathbf{R}_{1} $$ is unique where $\mathbf{Q}_{1}\in\mathbb{R}^{m\times n}$ has orthonormal columns and $\mathbf{R}_{1}$ is upper triangular with positive diagonal entries. Moreover, $\mathbf{G}=\mathbf{R}_{1}^{T}$ where $\mathbf{G}$ is the lower triangular Cholesky factor of $\mathbf{A}^{T}\mathbf{A}.$
Proof
Since $\mathbf{A}^{T}\mathbf{A} = \left(\mathbf{Q}_{1} \mathbf{R}_{1}\right)^{T} \left(\mathbf{Q}_{1} \mathbf{R}_{1}\right) = \mathbf{R}_{1}^{T}\mathbf{R}_{1} $ we see that $\mathbf{G}=\mathbf{R}_{1}^{T}$ is the Cholesky factor of $\mathbf{A}^{T}\mathbf{A}$. This factor is unique by Theorem 4.2.5. Since $\mathbf{Q}_{1} = \mathbf{A}\mathbf{R}_{1}^{-1}$ it follows that $\mathbf{Q}_{1}$ is also unique.