# Tikhonov regularization vs truncated SVD

To find $\mathbf{x}$ such that

$$A\mathbf{x}=\mathbf{b}$$

we can use least squares when the problem is not well posed. Further, we can use Tikhonov regularization when $A$ is ill-conditioned. In Tikhonov regularization, we minimize

$$\|A\mathbf{x}-\mathbf{b}\|_2^2+\|\Gamma\mathbf{x}\|_2^2$$

where $\Gamma$ determines the regularization properties. Alternatively, we could use the truncated SVD of $A$ to find the pseudoinverse. For SVD $A=U\Sigma V^T$, the truncated SVD is

$$U\Sigma_kV^T$$

where $\Sigma_k$ is composed of the first $k$ singular values. Truncating the SVD provides another means of regularization by producing solutions with smaller norms.

When is Tikhonov regularization similar (or even the same) as using the truncated SVD?

• Without loss of generality, we can take $U$ to be $m\times n$ while $\Sigma$ and $V$ are both square. Then the solution via Tikhonov regularization is $V(\Sigma^2 + V^T\Gamma^T\Gamma V)^{-1}\Sigma U^T b$ while the solution using the truncated SVD is $V\Sigma_k^+U^Tb$. The solutions are identical when $(\Sigma^2 + V^T\Gamma^T\Gamma V)^{-1}\Sigma = \Sigma_k^+$, which I believe is impossible if some of the discarded singular values are nonzero.
– user856
Dec 29, 2014 at 19:02

Consider the SVD of the $N\times N$ matrix A to be, $$A = U\Sigma V^T$$ where U and V are orthogonal matrices, and $\Sigma$ is a diagonal matrix with entries $$\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_N \geq 0$$ For a ill-conditioned matrix $A$, all the singular values decay gradually to zero and the condition number, i.e.

$cond(A)= \frac{\sigma_1}{\sigma_N}$ is very large.

The SVD approach (a.k.a spectral filtering) is to damp the effects caused by division by the small singular values. $$x_{naive} = A^{-1}b= V\Sigma^{-1}U^Tb = \sum_{i=1}^{N}\frac{u_i^Tb}{\sigma_i}v_i$$

The TSVD method is an example of the general class of methods that are called spectral filtering methods, which have form, $$x_{filt} = \sum_{i=1}^{N}\phi_i\frac{u_i^Tb}{\sigma_i}v_i$$ where the filter factors $\phi_i$ are chosen such that $\phi_i \approx 1$ for large singular values, and $\phi_i \approx 0$ for small singular values.

Now look at the filter equations,

$\bf{The~TSVD~Method}$ $$\phi_i = 1, i = 1,\cdots, k$$ $$= 0, otherwise$$ The parameter $k < N$ is called the truncation parameter.

$\bf{The~Tikhonov~Method}$ $$\phi_i = \frac{\sigma_i^2}{\sigma_i^2 + \alpha^2}, i = 1,\cdots, N$$ The parameter $\alpha > 0$ is called the regularization parameter. This choice of filter factors yields the solution vector $x_{\alpha}$ for the minimization problem, $min_x { ||b-Ax||^2_2 + \alpha^2 ||x||_2^2}$.

To answer your question, "when Tikhonov regularization becomes similar(or equal) to TSVD", we can see that as $\alpha \rightarrow 0$, $\phi_i \rightarrow 1$ which are the filter coefficients, and the Tikhonov method becomes similar to TSVD. You can think of this filtering as, TSVD uses a filter with a sharp jump from 0 to 1 and Tikhonov using a smoother approach (it does prevent oscillations to the solution.) For more details see Spectra and Filtering.