I'm trying to solve the following problem of a 'perturbed' QR factorization in preparation for a test of mine.
Q: Let $A \in \mathbb{C}^{n \times m}, m \geq n$ have a thin $QR$ factorization $A = QR$. Determine the $QR$ factorization of $A - Q e_1 e_1^* R$ in terms of the $QR$ factorization of $A$.
Attempted Solution: This new matrix has an interesting structure, that is: $$ Q e_1 e_1^* R = \begin{bmatrix} Q(:,1)R(1,1) & \dots & Q(:,1)R(1,n) \end{bmatrix} $$ where $Q(:,1)$ is referring to the first column of $Q$. This tells us that in fact: $$ A - Q e_1 e_1^* R = \begin{bmatrix} 0 & A(:,2) - Q(:,1)R(1,2) & \dots & A(:,n) - Q(:,1)R(1,n) \end{bmatrix} $$ So this perturbed matrix is $A$ with the first column zeroed out and with some '$QR$' pieces removed. However, I don't see an easy way to directly find the $QR$ factorization of this? Is there perhaps an argument I can make from the orthogonality of $Q(:,1)$ with respect to the rest of $Q$, and then perhaps there's a argument of how to augment the original $Q$ to triangularize this?
Any thoughts?
EDIT: I have discovered via running an example in octave, that $Q^*(A - Qe_1 e_1^* R)$ actually seems to produce an upper triangular matrix, will report back if this results in something interesting.
I'm foolish, this was actually easy.
Going off of my edit, we see that: \begin{align*} Q^*(A - Q e_1 e_1^* R) &= R - e_1 e_1^* R \end{align*} which is simply $R$ with its first row zeroed out, and critically is upper triangular. Thus $A - Qe_1 e_1^* R = Q (R - e_1 e_1^* R)$ is a QR factorization (which I could have alternatively seen by factoring a $Q$ out of the original matrix...)
Whoops.