Question about extending Tikhonov regularization

Question

I know that the Tikhonov regularization of a linear system has an analytical solution given by:

\begin{equation} \hat{\mathbf{x}} = \mathrm{arg\;min}\left( \left| \mathbf{Ax} - \mathbf{b} \right|^{2} + \left| \mathbf{\Gamma x} \right|^{2} \right) = \left( \mathbf{A}^{\top}\mathbf{A} + \mathbf{\Gamma}^{\top}\mathbf{\Gamma} \right)^{-1} \mathbf{A}^{\top} \mathbf{b} \end{equation}

Is it possible to extend this idea to obtain an analytical solution for a system with an additional weighting? i.e.

\begin{equation} \hat{\mathbf{x}} = \mathrm{arg\;min}\left( \left| \mathbf{Ax} - \mathbf{b} \right|^{2} + \left| \mathbf{\Gamma x} \right|^{2} + \left| \mathbf{U x} \right|^{2} \right) = \; ? \end{equation}

Thanks for any help!

Answer 1

Yes:

The solution to the Tikhonov regularization is derived by computing the first order optimality condition of functional to be minimized and then inverting the derived linear system. Applying the same process for your functional, let: $$ I(\mathbf{x}) = \left|\mathbf{Ax} - \mathbf{b}\right|^2 + \left|\mathbf{\Gamma x}\right|^2 + \left|\mathbf{Ux}\right|^2 $$ Then, taking the derivative w.r.t to $\mathbf{x}$ and setting this equal to zero, we get: $$ \frac{\partial I}{\partial\mathbf{x}}= 0 = 2 \left[\mathbf{A}^{T}\left(\mathbf{Ax}-\mathbf{b}\right) + \mathbf{\Gamma}^T\mathbf{\Gamma x} + \mathbf{U}^{T} \mathbf{Ux} \right] $$ Rearranging and solving for $\mathbf{x}$, gives: $$ \mathbf{x}=\left(\mathbf{A}^{T}\mathbf{A} + \mathbf{\Gamma}^T\mathbf{\Gamma} + \mathbf{U}^{T} \mathbf{U}\right)^{-1} \mathbf{A}^{T} \mathbf{b} $$