Convexity of $x\mapsto \mathrm{tr}(e^{-E\langle a,x\rangle}bb')$

Question

Let $E$ be a matrix with entries equal to one minus the identity matrix and $a,b$ known vectors. Is the function $$x\mapsto \mathrm{tr}(e^{-E\langle a,x\rangle}bb')$$

convex?

Answer 1

It is convex. Since $E$ admits an orthogonal diagonalisation $Q\operatorname{diag}(n-1,\,-1,\ldots,-1)Q^T$, if we put $c=Q^Tb$, the function in question can be rewritten as $$ x\mapsto\sum_{i=1}^n|c_i|^2\exp(-\lambda_i\langle a,x\rangle), $$ where $\lambda_1=n-1$ and $\lambda_2=\cdots=\lambda_n=-1$. This is a non-negatively weighted sum of convex functions. Hence it is convex.