= Davis-Kahan curvature lemma
{c}
{title2=$\frac\gamma2\|P-Q\|_F^2\leq\operatorname{tr}(H(P-Q))$}
If a <symmetric matrix> $H$ has largest <eigenvalue> $\lambda_1$ separated from $\lambda_2$ by $\gamma>0$, and $P=vv^\top$ is its leading rank-one <orthogonal projection matrix>, then every unit vector $u$, with $Q=uu^\top$, satisfies the displayed inequality. Expanding $u$ in an <orthonormal eigenbasis> gives $\lambda_1-u^\top Hu\geq\gamma(1-|u^\top v|^2)$, while $\|P-Q\|_F^2=2(1-|u^\top v|^2)$. The analogous statement for a smallest <eigenvalue> follows by replacing $H$ by $-H$.
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