I'm trying to understand the solution to Boyd and Vandenberghe Problem 5.30:
Boyd and Vandenberghe Problem 5.30
The Lagrangian is $$L(X,\nu)=\text{tr}X-\log\det X+\nu'\left(Xs-y\right),$$ so the stationary KKT condition is $$\nabla L(X,\nu)=I-X^{-1}+s\nu'=0,$$ but the solution implies $$\nabla L(X,\nu)=I-X^{-1}+\dfrac{\nu s'+s\nu'}{2}.$$ Did I compute the gradient incorrectly?
I think you have to use this property:
$$ \nu^TXs = \sum X_{ij} \nu_i s_j = tr(\nu s^T X) = tr(X^T (\nu s^T)^T) $$
X belongs to symmetric matrices $(X \in S^n)$ so: $X^T = X$ then:
$$ tr(Xs\nu^T) = tr(s\nu^TX) $$
we have obtained that $tr(\nu s^TX)=tr(s\nu^TX)$, consequently:
$$ tr(\nu s^TX) = [tr(\nu s^TX) + tr(s\nu^TX)]/2 = tr(\frac{\nu s^T + s\nu^T}{2}\cdot X) $$
and $\nabla_X$ of that is the term: $(\nu s^T + s\nu^T)/2$