Suppose we have the following optimization problem on $r\in \mathbb{R}$ with constraint on positive definite matrices $A,B$
\begin{eqnarray} \text{minimize } &\; r\\ \text{subject to } &\; A \prec r B \end{eqnarray}
What is a good way to show that solution is the largest generalized eigenvalue, using the definition of generalized eigenvalue from page 102 of Boyd & Vandenberghe's Convex Optimization:
It's important to note that by assumption of $A\in S^n$ and $B\in S^n_{++}$ (e.g. that $A$ is symmetric and $B$ is symmetric positive definite). Then
$$ A\prec r B \iff \sup_{\|u\|_2 =1 }u^T(A-rB)u < 0. \quad (*) $$ Define $\bar u$ by the generalized eigenvalue (the $u$ that achieves the supremum in $(*)$). Since $B$ is symmetric positive definite, then for any $u\neq 0$, $u^TBu> 0$, so $(*)$ is equivalent to $$ \frac{\bar u^TA\bar u}{ \bar u^TB\bar u } < r. $$ Since the problem is to minimize $r$, the optimal solution to the SDP is in fact $$ \frac{\bar u^TA\bar u}{ \bar u^TB\bar u } = r. $$
It is left to show that $$ \frac{\bar u^TA\bar u}{ \bar u^TB\bar u } = \sup_u \frac{u^TAu}{ u^TBu }=\lambda_{\max}(A,B) . $$ Assume that it is not true, e.g. there exists $\hat u$ where $$\frac{\hat u^TA\hat u}{\hat u^TB\hat u} - \xi= \frac{\bar u^TA\bar u}{ \bar u^TB\bar u } = r, \quad \xi > 0. $$ Then $$ \hat u^T(A-rB)\hat u = \xi \hat u^TB\hat u> 0 > \sup_{\|u\|_2=1}u^T(A-rB)u $$ which is not possible. Therefore $\lambda_{\max}(A,B) = r$.
