Can anyone explain this transformation to me :
$$ ||Ax-b||_2^2 + \delta ||x||_2^2 : A \in R^{m,n}, b \in R^m \rightarrow \\ \tilde{x} = (V^Tx, V_2^Tx), \tilde{b} = (U^Tb, U_2^Tb)\\ V_2 \in R^{n x (n-r)}, \, V_2^TV_2 = I, \, V_2^TV = 0\\ U_2 \in R^{m x (m-r)}, \, U_2^TU_2 = I, \, U_2^TU = 0\\ \rightarrow ||Ax-b||_2^2 + \delta ||x||_2^2 = \sum_{i=1}^r(\sigma_i \tilde{x}_i - \tilde{b}_i)^2 + \sum_{i=r+1}^m \tilde{b}_i^2, \, ||x||_2^2 = \sum_1^n\tilde{x}^2 $$
I get that a singular value decomposition is used :
$$A = U \text{diag} (\sigma) V^T = \sum_{i=1}^r \sigma_i u_i v_i$$
But I have no idea about the rest of the transform. I'd like to know the intuition behind it, if someone can point to a reference or explain, thank you.
Notice that for an orthogonal matrix, we have $\|Ux\|^2=\|x\|^2$, since rotation and reflection doesn't change the length.
\begin{align} \|Ax-b\|^2 + \delta \|x\|^2 &= \|U \Sigma V^Tx-b\|^2 + \delta \|x\|^2 \\ &= \|U (\Sigma V^Tx-U^Tb)\|^2 + \delta \|V^Tx\|^2 \\ &= \|(\Sigma (V^Tx)-U^Tb)\|^2 + \delta \|(V^Tx)\|^2 \\ &= \|(\Sigma \tilde{x}-\tilde{b})\|^2 + \delta \|\tilde{x}\|^2 \\ &=\sum_{i=1}^r (\sigma_i\tilde{x}_i-\tilde{b}_i)^2+\sum_{i=r+1}^m \tilde{b}_i^2 + \delta \|\tilde{x}\|^2\end{align}