I currently have this modified least squares problem:
$$\min_{x\in\mathbb R} \left \| Ax-b \right \|_{2}^{2} + \left \| Lx \right \|_{2}^{2}$$
which has a solution:
$$x = (A^{T}A + L^{T}L)^{-1} A^{T}b$$
I would like that given $L = \alpha I$ for some $α \neq 0$ to represent the solution explicitly in terms of α and the singular values and singular vectors of A.
The only way I can think of doing this is by writing A as its singular value decomposition:
$$A = U\Sigma V^{T}$$
From which I find that:
$$ \left \| Ax-b \right \|_{2}^{2} + \left \| Lx \right \|_{2}^{2} = \left \| U\Sigma V^{T}x - b \right \|_{2}^{2} + \left \| Lx \right \|_{2}^{2}$$
But I'm not sure how to arrive at a solution that also includes the singular values and singular vectors as well.
We have $$ A^tA + L^tL = V\Sigma^2 V^t + \alpha^2 I = V(\Sigma^2 + \alpha^2I) V^t $$ Let the diagonal matrix $\Gamma^2$ be $$ \Gamma^2 = \Sigma^2 + \alpha^2 I $$ Then $$ (A^tA + L^tL)^{-1} = V\Gamma^{-2}V^t $$ Thus, $x$ is given by $$ x=(V \Gamma^{-2} V^t)( V \Sigma U^t) b = V \Gamma^{-2} \Sigma U^t b . $$ Note that the $i$th diagonal value of $\Gamma^{-2}\Sigma$ is $\frac{\sigma_i}{\sigma_i^2+\alpha^2}$ and when $\alpha=0$, the solution reverts to the standard least-squares solution given by the SVD.