 Davis-Kahan curvature lemma

If a symmetric matrix $H$ has largest eigenvalue ${\lambda }_1$ separated from ${\lambda }_2$ by $\gamma >0$, and $P=v{v}^{\top }$ is its leading rank-one orthogonal projection matrix, then every unit vector $u$, with $Q=u{u}^{\top }$, satisfies the displayed inequality. Expanding $u$ in an orthonormal eigenbasis gives ${\lambda }_1-u^{\top }Hu\ge \gamma (1-|u^{\top }v{|}^2)$, while $\Vert P-Q{\Vert }_F^2=2(1-|u^{\top }v{|}^2)$. The analogous statement for a smallest eigenvalue follows by replacing $H$ by $-H$.