 Solution

Full column rank makes $B=A^{T}A$ positive definite: for nonzero $v$, $v^{T}Bv=\Vert Av{\Vert }^2>0$. Let $B=R^{T}R$ be its unique upper-triangular Cholesky decomposition with positive diagonal. Defining $Q=A{R}^{-1}$ givesThus the skinny QR decomposition exists. If $A=\tilde{Q}\tilde{R}$ is another such decomposition, orthonormality gives $A^{T}A={\tilde{R}}^T\tilde{R}$. Cholesky uniqueness forces $\tilde{R}=R$, and then $\tilde{Q}=A{R}^{-1}=Q$. Therefore the skinny factorization with positive diagonal is unique. This is the thin QR factorization from Cholesky decomposition; no inverse of the rectangular $Q$ is assumed.