 Kitaev geometrical lemma

Suppose positive operators $A,B$ have positive eigenvalues at least $\gamma$ and their nullspaces meet only at zero. If $\vartheta$ is their smallest angle between two subspaces, then $A+B\ge 2\gamma {\sin }^2(\vartheta /2)I$. Indeed $A+B\ge \gamma (2I-P-Q)$, while $(P+Q{)}^2\le (1+\cos \vartheta )(P+Q)$ gives $P+Q\le (1+\cos \vartheta )I$. With a common nullspace, apply the same proof on its orthogonal complement to bound the positive spectral gap.