= Kitaev geometrical lemma
{c}
{title2=$\lambda_{\min}(A+B)\geq2\gamma\sin^2(\vartheta/2)$}
= Kitaev's geometrical lemma
{c}
{synonym}
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 $\cos\vartheta=\|PQ\|$. Then
$$
A+B\geq\gamma(1-\cos\vartheta)I=2\gamma\sin^2(\vartheta/2)I.
$$
When 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, $A\geq\gamma(I-P)$ and $B\geq\gamma(I-Q)$. For $p\in\operatorname{ran}P$, $q\in\operatorname{ran}Q$, $|\langle p,q\rangle|\leq\|PQ\|\|p\|\|q\|$. The map $(p,q)\mapsto p+q$ consequently has squared <operator norm> at most $1+\|PQ\|$, so $P+Q\leq(1+\|PQ\|)I$. 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 $A\geq\gamma(I-P)$ and $B\geq\gamma(I-Q)$. An assertion only about eigenvalues can be vacuous in the presence of continuous <spectrum>, and does not ensure this uniform gap.
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