# Slater's Condition for Dual Problem

I am reading the paper "A Semidefinite Optimization Approach to Quadratic Fractional Optimization with a Strictly Constraints" by Maziar Salahi & Saeed Fallahi.

I this paper, they tried to prove that if the Slater condition holds for the primal SDP problem: \begin{equation*} \begin{aligned} \min_{X} \mathrm{tr}(C^\top X)\\ \text{s.t.}~\mathrm{tr}(A^\top X) = 1\\ \mathrm{tr}(B^\top X) \preceq 0 \\ X \succeq 0 \end{aligned} \end{equation*} as well as its dual problem: \begin{equation*} \begin{aligned} \max_{\lambda,\eta} \eta\\ \text{s.t.}~C^\top -\eta A^\top + \lambda B^\top = Z\\ \lambda\geq 0 \\ Z \succeq 0 \end{aligned} \end{equation*} then both problems attain their optimal values and the duality gap is zero.

Because trace is a linear function, in order to show the strong duality, I know one has to show the Slater condition holds for the primal problem.

My Question is: why should I also have to show the Slater condition holds for the dual problem?

If either the primal or the dual satisfies Slater's condition, strong duality holds. However, the problem for which Slater's condition holds can still be unbounded, so you cannot conclude that "both problems attain their optimal values". To show that the primal problem is bounded you could give a feasible point for the dual or vice versa. That is not the only way. Instead, you can conclude that the primal is bounded if $$C$$ is positive semidefinite.