Let $S\subset\mathbb{R}$ be a convex set and $\mathbb{S}^{n}$ be the set of real symmetric matrices of order $n\times n$.

A matrix valued function $\Gamma: S \rightarrow \mathbb{S}^{n}$ is said to be convex if for all $x_1,x_2 \in S$ and for all $\lambda \in (0,1)$ one has $$\Gamma\left(\lambda x_{1}+\left(1-\lambda\right)x_{2}\right)\preceq\lambda\Gamma\left(x_{1}\right)+\left(1-\lambda\right)\Gamma\left(x_{2}\right),$$ where $\prec$ denotes the Loewner partial order, i.e. $A \prec B$ if $A - B$ is negative definite.

Is the matrix valued function $f: [0,c] \rightarrow \mathbb{S}^{n}$ given by $$ f(t) = e^{At} e^{A^{T} t} $$ convex for any $A \in \mathbb{R}^{n\times n}$ and $c>0$?

Numerical experiments suggest that this statement is true, but I could not prove.

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    $\begingroup$ Although the original claim is false (Mike Jury has an excellent explicit counterexample), a much weaker related claim is true. Claim. Let $A \in M_n(\mathbb{C})$. The function $f: [0,c] \to \mathbb{R}_+$ defined by \begin{equation*} f(t) = \|e^{tA}\| \end{equation*} is convex for any unitarily invariant norm $\|\cdot\|$. $\endgroup$ – Suvrit May 13 '15 at 18:01
  • $\begingroup$ Thanks Suvrit, this claim is also interesting to know :) $\endgroup$ – Shamisen May 13 '15 at 18:58
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    $\begingroup$ For completeness I suppose it's worth noting that the claim is true if $A$ is normal. $\endgroup$ – Noah Stein May 13 '15 at 20:11
  • $\begingroup$ @NoahStein Thanks for the comment. Now I'm wondering if $A$ diagonalizable could imply the claim... $\endgroup$ – Shamisen May 13 '15 at 21:54
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    $\begingroup$ Actually, on further thought there is a simple necessary and sufficient condition for $f$ to be convex, I will edit my answer to describe it. $\endgroup$ – Mike Jury May 13 '15 at 23:58

I think the $3\times 3$ Jordan block $$ A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} $$ is a counterexample.

For this $A$, we have $$ \exp{At}=\begin{pmatrix} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1\end{pmatrix} $$ so $$ f(t) = \begin{pmatrix} 1+t^2+\frac{t^4}{4} & t+\frac{t^3}{2} & \frac{t^2}{2} \\ t+\frac{t^3}{2} & 1+t^2 & t \\ \frac{t^2}{2} & t & 1\end{pmatrix} $$ Take any $c>0$. The last column of $\frac{1}{2}(f(c)+f(0)) - f(\frac{c}{2})$ is $$ \begin{pmatrix} \frac{c^2}{8} \\ 0 \\ 0\end{pmatrix} $$ so the difference $\frac{1}{2}(f(c)+f(0)) - f(\frac{c}{2})$ cannot be positive for any $c$.

EDIT: In fact it is possible to obtain a necessary and sufficient condition for the convexity of $f$ in terms of $A$. Namely, I claim that if the matrix $$ A^2 +A^{\intercal 2}+ 2A A^\intercal $$ is positive semidefinite, then $f$ is convex on $\mathbb R$, and if not, then $f$ is not convex on any interval $[a,b]$. To see this, one can first check that $f(t)$ is convex on $[a,b]$ if and only if the scalar functions $$ f_u(t):=\langle f(t)u, u\rangle $$ are convex on this interval for all $u\in\mathbb R^n$. But since the functions $f_u$ are smooth, they will be convex if and only if the second derivatives are nonnegative. Computing we find $$ f_u^{\prime\prime}(t) = \langle (A^2 +A^{\intercal 2} + 2AA^\intercal)\exp(tA^\intercal)u, \exp(tA^\intercal)u\rangle $$ and the claim follows.

  • $\begingroup$ Excellent! Thanks for the clear answer! :) $\endgroup$ – Shamisen May 13 '15 at 18:59

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