Kitaev geometrical lemma (${\lambda }_{\min }(A+B)\ge 2\gamma {\sin }^2(\vartheta /2)$)

Let $A,B$ be positive operators on a finite-dimensional Hilbert space, with every nonzero eigenvalue at least $\gamma >0$. Let $P,Q$ be the orthogonal projections onto $\ker A,\ker B$ and define $\cos \vartheta =\Vert PQ\Vert$. ThenWhen the kernels have no common nonzero vector, this gives a positive lower bound. The projection formula also covers a zero kernel without taking a supremum over an empty unit sphere.

Indeed, $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. For $p\in \mathrm{ran}P$, $q\in \mathrm{ran}Q$, $|⟨p,q⟩|\le \Vert PQ\Vert \Vert p\Vert \Vert q\Vert$. The map $(p,q)\mapsto p+q$ consequently has squared operator norm at most $1+\Vert PQ\Vert$, so $P+Q\le (1+\Vert PQ\Vert )I$. Combining these inequalities proves the result. With common kernel $N$, restrict to $N^{\perp }$ and use the kernel projections there to bound the positive spectral gap.

In infinite dimension, replace the eigenvalue condition by the actual operator inequalities $A\ge \gamma (I-P)$ and $B\ge \gamma (I-Q)$. An assertion only about eigenvalues can be vacuous in the presence of continuous spectrum, and does not ensure this uniform gap.