Is the scalar field ${\bf A} \mapsto \mbox{Tr} \left( {\bf A} {\bf B} {\bf A}^\top {\bf C} \right)$ convex?

Question

Let ${\bf A}, {\bf B}, {\bf C}$ be $d \times d$ (symmetric) positive semidefinite matrices. Let $\mbox{Tr}$ denote the trace operator. Is the scalar field ${\bf A} \mapsto \mbox{Tr} \left( {\bf A} {\bf B} {\bf A}^\top {\bf C} \right)$ convex?

I guess it is convex, but could not prove it formally. Is there any idea or reference that I can take a look at?

Have you tried using the square roots of $B$ and $C$ and the cyclic property of the trace to write the trace of the product as the Frobenius norm of a matrix? — Rodrigo de Azevedo, Dec 10 '23 at 09:27
@RodrigodeAzevedo Thank you very much! Substituting the square roots of B and C shows the convexity. — sabo, Dec 11 '23 at 17:48

score 3 · Accepted Answer · answered Jan 30 '24 at 20:58

Since $\boldsymbol{B}, \boldsymbol{A} \in \mathcal{S}_{+}^{n}$ then they can be written as:

$$ \boldsymbol{B} = \boldsymbol{L} \boldsymbol{L}^{T}, \; \boldsymbol{C} = \boldsymbol{R}^{T} \boldsymbol{R} $$

Then it becomes:

$$ \operatorname{Tr} \left( \boldsymbol{A} \boldsymbol{B} \boldsymbol{A}^{T} \boldsymbol{C} \right) = \operatorname{Tr} \left( \boldsymbol{A} \boldsymbol{L} \boldsymbol{L}^{T} \boldsymbol{A}^{T} \boldsymbol{R}^{T} \boldsymbol{R} \right) = \operatorname{Tr} \left( \boldsymbol{R} \boldsymbol{A} \boldsymbol{L} \boldsymbol{L}^{T} \boldsymbol{A}^{T} \boldsymbol{R}^{T} \right) = \operatorname{Tr} \left( \boldsymbol{R} \boldsymbol{A} \boldsymbol{L} \left( \boldsymbol{R} \boldsymbol{A} \boldsymbol{L} \right)^{T} \right) = {\left\| \boldsymbol{R} \boldsymbol{A} \boldsymbol{L} \right\|}_{F}^{2} $$

Now, $\boldsymbol{R} \boldsymbol{A} \boldsymbol{L}$ is a linear function of $\boldsymbol{A}$ hence the squared Frobenious norm is a convex function of $\boldsymbol{A}$.

Is the scalar field ${\bf A} \mapsto \mbox{Tr} \left( {\bf A} {\bf B} {\bf A}^\top {\bf C} \right)$ convex?

1 Answers1