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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.

KingLogic
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Akahs
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1 Answers1

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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.

user8675309
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  • So basically it's not unique because of the ambiguity of the sign of the last column of Q? Then I think it's hard to make a clear cut -- even for full rank matrices we ignore the sign ambiguity by assuming positive diagonals in R. But we have a zero here so, I guess yeah... – Akahs Oct 13 '20 at 04:19
  • incidentally, same idea with Polar Decomposition https://math.stackexchange.com/questions/3834754/how-to-show-that-the-unitary-matrix-in-a-polar-value-decomposition-a-up-is-uni/3834827 . I generally wouldn't expect uniqueness when dealing with singular matrices. – user8675309 Oct 13 '20 at 06:44