Say if I have a matrix $A$ whose first $n-1$ columns are linearly independent, and the nth column is a linear combination of the previous $n-1$ columns. Is the QR decomposition also uniquely determined in this case? Here's my reasoning:
If we look at the first $n-1$ columns of $Q$ and $R$, they are definitely unique, but then the last column of $Q$ is also uniquely defined. Then the last column of $R$ must have $0$ as the last element and the previous $n-1$ elements are defined by the linear combination. So in summary everything seems to be uniquely defined.
I assume $A$ is square and you are doing the standard $QR$ factorization associated with square matrices -- i.e. $Q$ and $R$ are both square. (The fact that there are ways to do this without $Q$ and $R$ both being square is a red flag that this isn't going to be unique).
Let $D$ be a diagonal matrix with all ones on the diagonal, except the bottom right entry is $-1$.
$A=QR = Q\big(I\big)R =Q\big(DD\big)R = \big(QD\big)\big(DR\big)=\big(QD\big)R$
But $Q\neq QD$ so it is not unique.