1
$\begingroup$

Given a $2 \times 2$ matrix, $C$, suppose I want to compute a filter matrix $H = C^{-1}$ and that I need to add regularization for practical purposes (e.g., for an audio filter, regularization is needed to limit the gain of the filter at certain frequencies where the condition number of $C$ is large.) The linear system of equations being solved is $CH = I$, where $I$ is the $2 \times 2$ identity matrix.

My approach would be to use Tikhonov regularization where the solution (at each frequency) is given by $H = \left(C^* C + \beta I\right)^{-1} C^*I$, where $\beta$ is a scalar regularization parameter and * denotes taking the complex conjugate, since this minimizes the objective function $\lVert CH-I \rVert^2 + \beta\lVert H \rVert^2$. So, with an appropriately chosen $\beta$, we can essentially constrain the filter gain (since $\lVert H \rVert$ effectively corresponds to the max. gain of $H$), albeit at the expense of $H$ satisfying $CH = I$ less optimally compared to using $\beta = 0$.

Another approach might be to compute $H$ as $H = \displaystyle\frac{(\det C)^*}{|\det C|^2+\beta}\text{adj}(C)$, where $\text{adj}(C)$ corresponds to the adjugate of $C$ and $\det C$ to its determinant. Note that, for $\beta = 0$, this corresponds precisely to $C^{-1}$. This is an approach I came across recently but have not seen before.

I'm pretty sure the two solutions are not equivalent, but if they are, then a proof of the same would be appreciated. Assuming they are not, I've been trying to determine the benefits of adopting one approach over the other. For one, the latter can only be done if $C$ is square (and, from a computation standpoint, not too large since it requires explicit computation of a determinant and the adjugate). Also, $\beta$ enters the solution differently and so optimizing the value of $\beta$ might be easier in one method than the other. Apart from that, I don't have much insight regarding the differences between the two since I'm not really familiar with the latter approach.

As a response from the community, I'm looking for confirmation as to whether or not the two approaches are different, and for comments on when one might want to use one approach over the other, including a more intuitive explanation of the latter approach, if possible. Links to papers/articles/books that use the latter approach would count and be appreciated. Thanks!

$\endgroup$
0
$\begingroup$

This alternative approach seems fishy to me. Consider the two-dimensional case with $C$ having eigenvalues $\lambda_0, \lambda_1$ that are real and positive and $\lambda_0 \ll \lambda_1$. You have $$ \det C = \lambda_0\cdot \lambda_1 $$ What you want to prevent with the regularisation is that your inverse gets a too big eigenvalue $\lambda_0^{-1}$, i.e. you want to cap that to at most $\beta^{-1}$. Meanwhile, $\lambda_1^{-1}$ is fine. But if you put in the corresponding eigenvector $v_1$ into that alternative formula, you get $$\begin{align} Hv_1 =& \frac{\det C}{|\det C|^2 + \beta}\cdot\operatorname{adj}C\: v_1 \\ =& \frac{\lambda_0^2\cdot\lambda_1^2}{\lambda_0^2\cdot\lambda_1^2 + \beta}\cdot C^{-1}\: v_1 \\ =& \frac{\lambda_0^2\cdot\lambda_1}{\lambda_0^2\cdot\lambda_1^2 + \beta}\cdot v_1 \\ \approx& \frac{\lambda_0^2\cdot\lambda_1}{\beta}\cdot v_1 \end{align}$$ I.e. the scaling for all inputs depends on $\beta$, not just for the low-eigenvalue ones.

I suppose this formula might still work in some situations (perhaps when the determinant is mostly imaginary?), but it certainly isn't appropriate for the general case.

Use Tikhonov.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.