Let $A = \begin{bmatrix} R & w\\ 0 & v \\ \end{bmatrix}$ and $b = \begin{bmatrix} c\\ d\\ \end{bmatrix}$,
where $R$ is a $k × k$ block $c,w ∈ R^k, v, d ∈ R^{m−k},$ and the zero is a block of appropriate dimension. If $A ∈ R^{m×(k+1)}$ has full column rank, then show that$$min_x ||Ax-b||_2^2 = ||d||_2^2 - (\frac{v^Td}{||v||_2})^2$$
Hint: If we write $x = (\tilde x,\alpha)$ where $\tilde x \in \Bbb R^k$ and $\alpha \in \Bbb R$, then we have $$ \|Ax - b\|_2^2 = \|R\tilde x - (c - \alpha w)\|_2^2 + \|\alpha v - d\|_2^2. $$ Note also that because $R$ is square with linearly independent columns, it is invertible.