I have an overdetermined system
$$Ax=b$$
where $A \in \mathbb{R}^{m \times n}$ has full column rank and $b \in \mathbb{R}^{m}$. Let $\hat{x}$ denote the least squares solution.
If am given $c \in \mathbb{R}^n$, how can I compute $\alpha = c^T \hat{x}$ without solving for $\hat{x}$ explicitly?
I am given the hint that if I consider $Z$ as a Householder matrix such that $Z^T c$ is a multiple of $I_n(:,n)$, it follows that $\alpha = (Z^Tc)^T \hat{y}$, where $\hat{y}$ is the least squares solution to $\|\tilde{A}y-b\|_2$, with $y=Z^Tx$ and $\tilde{A}=AZ$
In looking at $\alpha = (Z^Tc)^T$, I concluded (not sure if 100% correctly) that $\alpha$ is the product of 2-norm of $c$ and last element of $\hat{y}$: i.e., $\alpha = \|c\|_2\cdot \hat{y}[n] \cdot \mbox{sign}(c[n])$, but I am not sure how to proceed.