Affine function $f(\lambda)=a^T\lambda + a_0$ where $a, \lambda\in \mathbb{R}^n,a_0\in \mathbb{R}$ and $\lambda$ is in a unit simplex,i.e., $\sum\limits_{i=1}^n \lambda =1, \lambda\in \mathbb{R}^n_+$.
Show that this affine function is equivalen to $\lambda^T(\tfrac{1}{2}(ea^T + ae^T) + a_0 E)\lambda$, where $E= ee^T$ and $e$ is a vector with components all valued $1$.
My intutition is as follows.
Exploit the fact that $\lambda^Te=e^T\lambda =1 $ by left multiplying $\lambda^Te$ to the function $f(\lambda)$, and then just right multipying $e^T\lambda$ to the single term $\lambda^Tea_0$.
But the result is
$\lambda^T(ea^T + a_0 E)\lambda$... what is wrong with my deduction ?
Just go in the reverse direction. $$\lambda^T(\tfrac{1}{2}(ea^T + ae^T) + a_0 E)\lambda = \tfrac{1}{2} \lambda^Tea^T\lambda + \tfrac{1}{2}\lambda^Tae^T\lambda + a_0\lambda^Tee^T\lambda$$ The first term: $$\lambda^Tea^T\lambda=(\lambda^Te)(a^T\lambda)=a^T\lambda$$ The second term: $$\lambda^Tae^T\lambda=(\lambda^Ta)(e^T\lambda)=\lambda^Ta = a^T\lambda$$ The third term: is the transpose of the first, so it's the same. And finally, $$\lambda^Tee^T\lambda=(\lambda^Te)(e^T\lambda)=1.$$ So the expression equals $$\tfrac{1}{2} a^T \lambda + \tfrac{1}{2} \lambda^Ta + a_0 = a^T\lambda + a_0$$ To finish your approach, just note that for a real scalar $b$, $b=\tfrac{1}{2}(b+b^T)$. So $$ \begin{aligned} \lambda^T(ea^T+a_0E)\lambda &= \tfrac{1}{2}\lambda^T(ea^T+a_0E)\lambda + \tfrac{1}{2}\left(\lambda^T(ea^T+a_0E)\lambda\right))^T \\ &= \tfrac{1}{2}\lambda^T(ea^T+a_0E)\lambda + \tfrac{1}{2}\lambda^T(a^Te+a_0E)\lambda \\ &= \tfrac{1}{2}\lambda^T(ea^T+ae^T+2a_0E)\lambda = \lambda^T(\tfrac{1}{2}(ea^T+ae^T)+a_0E)\lambda.\end{aligned}$$