Given image $x$ and a transformation (blurring) $K$, we get a blurry image $f$. The blurring transformation $K$ is ill conditioned. For a given $f$, find $x$.
Solution via regularization: consider $\alpha>0$ and solve $\min\{||Kx-f||^2+\alpha^2||x||^2\}$. Compare to $\min\left\{ \left|\left|\begin{bmatrix}f\\0\end{bmatrix}-\begin{bmatrix}K\\ \alpha I\end{bmatrix}x\right|\right|^2 \right\}$
By solving $\min\{||Kx-f||^2+\alpha^2||x||^2\}$, I have $x=(K^TK+\alpha^2I)^{-1}K^Tf$, solving $\min\left\{ \left|\left|\begin{bmatrix}f\\0\end{bmatrix}-\begin{bmatrix}K\\ \alpha I\end{bmatrix}x\right|\right|^2 \right\}$, I get $x=(K^TK+\alpha^2I)^{-1}K^Tf$ as well.
Is this coincident or I did something wrong? If this isn't an coincident, I don't quite see the connection between these functions? If no, can someone tell me what should the answer like? Thanks
Recall: $$ \left| \left| \begin{bmatrix}\vec{a} \\ \vec{b}\end{bmatrix} \right|\right|_2^2 = \sum_i a_i^2 + \sum_jb_j^2 = ||\vec{a}||_2^2 + ||\vec{b}||_2^2 $$ Thus: \begin{align} \left| \left| \begin{bmatrix}f \\ 0\end{bmatrix} - \begin{bmatrix}K \\ \alpha I\end{bmatrix}x \right|\right|_2^2 &= \left| \left| \begin{bmatrix}f \\ 0\end{bmatrix} - \begin{bmatrix}Kx \\ \alpha x\end{bmatrix} \right|\right|_2^2 \\[1mm] &= \left| \left| \begin{bmatrix}f - Kx \\ -\alpha x\end{bmatrix} \right|\right|_2^2 \\[2mm] &=||f-Kx||_2^2 +||{-}\alpha x||_2^2 \\ &=||Kx-f||_2^2 +\alpha^2||x||_2^2 \\ \end{align}