# Maximum Singular Value of $(A \cos\phi+B \sin\phi) \in \mathbb{R}^{2 \times 2}$

Problem

I am wondering if there is a way to efficiently compute the maximum singular value of \begin{align} C(\phi)=A \cos\phi+B \sin\phi, \end{align} where A, B and C are real 2x2 matrices.

Numerical Approach (edited)

So far, I could only solve the problem 'numerically' by

• Direct optimization (e.g. using matlabs's fmincon), which is sensitive to the choice of the intial value of $\phi$. Other (global) optimizers are more reliable but need far more evaluations. Maybe an appropriate bisection algorithm would help here? Suggestions?
• Rank-1 tensor approximation (e.g. using 'tenser toolbox' for matlab), which also needs quite some iterations. I am not very familiar with tensor algebra so I am not sure if this approach is even valid in general.

Notes (edited)

• It can be assumed that either $A$ or $B$ is positive semidefinite.
• QZ-decompositon may be applied to make $A$ and $B$ (and thus also $C$) triangular. But I am not sure if this helps.

In the special case of $\phi=\pi/4$, you're essentially asking for the maximum singular value of $A+B$. So without further structural assumptions imposed on $A$ and $B$, the question is basically unanswerable, as the eigenvalues of $A+B$ can basically be anything within the spectrum of $A$ and $B$.
• Thanks for your answer. If it helps, it can be assumed that either A or B is positive semidefinite. Furthermore, QZ-decompositon may be applied to make A and B (and thus also C) triangular . So far, I could solve the problem numerically by a) direct optimization (e.g. using matlabs's fmincon) which is sensitive to the choice of the intial value of $\phi$. Other (global) optimizers are more reliable but need far more evaluations. An appropriate bisection method may help?! OR b) rank-1 tensor approximation (e.g. using tenser toolbox for matlab), which also needs quite some iterations. – puman Jul 3 '17 at 11:03
• The additional constraint makes no difference. You can easily come up with $A$ and $B$ that satisfy all your assumptions, yet $A+B$ can be literally anything. – Yining Wang Jul 8 '17 at 1:39