Is $X \mapsto \operatorname{tr} (A e^{X})$ convex?

Question

The function $X \mapsto \operatorname{tr} e^X$ is convex (because it's a spectral function: $\sum \exp(\lambda_i)$ with $\lambda_i$ the eigenvalues of $X$. I'm pretty sure that $X \to \operatorname{tr} (A e^X)$ inherits from the same property when $A$ is Hermitian positive definite (maybe only over the set of matrices $X$ that are Hermitian) but I don't know what are the arguments?

Thanks

Answer 1

Although we already know that the conjecture is false, let us try to seek for a more systematic explanation.

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Your conjecture is equivalent to the claim that all of diagonal entries

$$ X \mapsto [e^X]_{kk}, \qquad 1 \leq k \leq n $$

are convex on the space of $n\times n$ Hermitian matrices. Here, $[B]_{kk}$ denotes the $(k, k)$-entry of the matrix $B$.

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Now let us focus on the case where $X$ ranges over $2\times 2$ symmetric real matrices. Write

$$ X = \begin{pmatrix} a & b \\ b & c \end{pmatrix} $$

and define two functions $s = s(X)$ and $q = q(X)$ by

$$s(X) = \frac{a+c}{2}, \quad q(X) = \frac{1}{2}\sqrt{\smash[b]{4b^2 + (a-c)^2}}. $$

Then the following general formula holds:

$$ e^X = e^s \left( \cosh q I + \frac{\sinh q}{q} (X - sI) \right). $$

From this, we have an explicit formula for the $(1,1)$-entry of $e^X$

$$ [e^X]_{11} = e^s \left( \cosh q + \frac{a-c}{2}\cdot\frac{\sinh q}{q} \right) $$

and we can test whether this function is convex or not.

This function almost looks like a convex function because all the functions $X \mapsto e^s$, $X \mapsto \cosh q$ and $X \mapsto \sinh q/ q$ are convex and positive. On the other hand, since they are combined in a rather arbitrary way, we may expect that convexity may break down at some point.

Indeed, The figure below is a graph of the Hessian determinant of the function

$$ (b, c) \mapsto \bigg[\exp\begin{pmatrix} 0 & b \\ b & c \end{pmatrix} \bigg]_{11} $$

$\hspace{6em}$

which shows that $X \mapsto [e^X]_{11}$ cannot be convex.

Answer 2

After generating several matrices, I've finally found a numerical counterexample.

Consider $A = \begin{pmatrix} 5.1056 & -2.0071\\ -2.0071 & 1.1172 \\ \end{pmatrix}$, $X_0 = \begin{pmatrix}-2.9808 & -1.8849\\ -1.8849 & -1.4978\\ \end{pmatrix}$ and $ D = \begin{pmatrix} -0.4262 & 0.8774 \\ 0.8774 & -2.2114 \end{pmatrix} $. Consider $f : t \in \mathbb{R} \mapsto \mathrm{tr}(A e^{X_0 + t D})$. Note that $A$ is sdp and $X_0$, $D$ and $X_0 + t D$ are hermitians. Plot $f(t)$ for $t \in [-40, 1.5]$. And you get the curve below, which is not convex. Hence $X \mapsto \mathrm{tr}\;A e^X$ is non convex.