Convex objective function of matrix with prescribed determinant and trace

Question

I have real symmetric positive definite matrix $M = \left(\begin{matrix} a & b \\ b & c \end{matrix}\right)$ where $a,b,c \in R,\ a,c>0,\ \left|b\right|<2\sqrt{a c}$.

I want to define CONVEX objective function of $M$ (preferably polynomial) which has minimum value for prescribed value of trace and determinant of $M$, i.e. $tr(M) = L$ and $det(M) = V$.

Natural definition of such function is probably

$f(M) = f(a,b,c) = \left( \frac{tr(M)}{L} - 1 \right)^2 + \left( \frac{det(M)}{V} - 1 \right)^2 = \left( \frac{a + c}{L} - 1 \right)^2 + \left( \frac{a c - b^2}{V} - 1 \right)^2$

The problem is that this function is not convex.

Does anybody have an idea how to define this function to be convex or can it be proved that this is not possible?

Thank you for any help.

Answer 1

You only need a convex function that involves the determinant. You can then minimize it subject to the constraint $\text{tr}\; M = L$ which is a linear constraint and, therefore, easy to handle.

That said, I believe that no such function exists so that $\phi(M)$ is minimal where $\text{det}\; M=V$ and only there, in general. Take, for example, the case where you want $V=0$ and shift your function so that $\phi(M)=0$ for all matrices with zero determinant. Then you are looking for a function that is zero for all degenerate matrices and positive for all other matrices. Now consider $M_1,M_2$ so that $\text{det}\; M_i=0$, i.e., $\phi(M_i)=0$. Convexity of $\phi$ would imply that $\phi((M_1+M_2)/2)=0$, but even if $\text{det}\;M_i=0$ you typically do not have $\text{det}\;(M_1+M_2)/2=0$ -- in other words, the set of matrices where $\phi$ is minimal (the set of matrices with zero determinant) is not convex itself, and consequently the objective function could only be convex if it is zero not only on the degenerate matrices but indeed their convex hull. But then, the objective function would also be zero on some matrices that do not have zero determinant -- not what you probably want.