Intuition for Kitaev's geometrical lemma Background
Kitaev's geometrical lemma is a linear algebraic result which is very useful within Hamiltonian complexity theory. Under certain conditions, it allows one to bound the spectrum of the sum of two linear operators with a formula related to the geometry of their kernels. It gets its name from the fact that it was used by Kitaev to prove that the local Hamiltonian problem is QMA-hard.
To state it formally, let me first establish some notation.
Notation
Given a linear operator $L$ on some finite-dimensional vector space $W$, we say $L \geq 0$ if all of $L$'s eigenvalues are greater than or equal to 0. Similarly $L \geq c$ for some constant $c$ means that $L - c I \geq 0$, where $I$ is the identity operator.
Let $A, B \geq 0$. Denote the minimum nonzero eigenvalues of $A$ and $B$ as $\lambda_{\mathrm{min}} (A |_{\mathrm{supp} A})$ and  $\lambda_{\mathrm{min}} (B |_{\mathrm{supp} B})$ respectively.
The geometrical content of Kitaev's geometrical lemma comes from the notion of an angle between subspaces. For two subspaces $U$ and $V$ of a vector space $W$, we define the angle $\theta(U, V)$ between $U$ and $V$ via
$$
\cos \theta(U,V) = \max_{\mathbf{u} \in U, \mathbf{v} \in V} \left| \langle \mathbf{u} , \mathbf{v} \rangle \right|,
$$
where the maximization is performed over all unit vectors in $U$ and $V$.
Formal statement
Kitaev's geometrical lemma can then be stated as follows.

Lemma: Let $A$ and $B$ be positive semi-definite linear operators on some finite dimensional vector space $W$. Then
$$
A + B \geq \min\left(\lambda_{\mathrm{min}} (A |_{\mathrm{supp} A}), \lambda_{\mathrm{min}} (B |_{\mathrm{supp} B})\right) \cdot (1 - \cos \theta ),
$$
  where $\theta = \theta(\ker A, \ker B)$ is the angle between the kernels of $A$ and $B$ in $W$.

See Ref. 1 for a proof.
Question
Is there some intuition for this lemma? There is an algebraic proof in Ref. 1 which I can understand, but it would be nice to see whether there is a geometric proof, or at least something that transparently connects the spectrum of these operators to the geometry of their kernels.
References
Ref. 1: A. Kitaev, A. Shen and M. Vyalyi, Classical and Quantum Computation, American Mathematical Society, 2002.
 A: Kitaev's lemma isn't about relating the spectra of $A$ and $B$ to the geometry of their kernels. More deeply, it's really about the eigenvalues of sums of projection operators. Indeed, the only eigenvalue of $A$ or $B$ that comes in to play in Kitaev's theorem is the smallest nonzero eigenvalue of $A$ or $B$.
Largest eigenvalue of the sum of projections.
Let $W$ be a finite-dimensional complex inner product space, and let $U,V\subseteq W$ be subspaces. Denote the projection operator onto the subspaces $U$ and $V$ as $\Pi_U$ and $\Pi_V$ respectively.
First note that we can always find unit vectors $u\in U$ and $v\in V$ such that $\langle u,v\rangle =\cos\theta$, where $\theta=\theta(U,V)$ is the angle between the the subspaces $U$ and $V$. These vectors are the two vectors in $U$ and $V$ that are the "closest" to each other in the sense that they have the smallest angle between them among all possible choices of unit vectors in $U$ and $V$. We make a few further remarks before moving on to Kitaev's lemma. 
For these vectors $u$ and $v$, it holds that
$$
\langle v,\Pi_Uv \rangle =\cos^2 \theta =\langle u,\Pi_V u\rangle.
$$
Indeed, if it were otherwise the case that $\langle v,\Pi_Uv \rangle>\cos^2\theta=\langle u,v\rangle^2$, then $v$ would not be optimal for computing the angle $\theta$. A similar argument holds for $u$.
Now suppose we wanted to find the unit vector $w\in W$ that maximizes the quantity
$$
\langle w,\, (\Pi_U+\Pi_V)w\rangle.\tag{1}
$$
Geometrically, we should choose $w$ to be the vector that is as close to both $u$ and $v$ as possible (in the sense that the angles between $u$, $v$, and $w$ are as small as possible), since $u$ and $v$ are already the vectors in $U$ and $V$ that are as close to each other as possible. 

As we see from the above figure, the vector $w$ that minimizes the quantity in (1) will be the angle bisector between $u$ and $v$. That is, we should choose $w$ to be the bisector
$$
w = \alpha(u+v),
$$
where $\alpha = \frac{1}{\sqrt{2(1+\cos\theta)}}$ is the normalizing factor such that $\lVert w\rVert =1$, in order to minimize (1).
Relation to Kitaev's lemma
We now relate the above observations to Kitaev's lemma. For clarity, we define the quantity $\lambda =\min\{\lambda_{\mathrm{min}} (A |_{\mathrm{supp} A}), \lambda_{\mathrm{min}} (B |_{\mathrm{supp} B})\}  $. Kitaev's lemma can be restated as follows:

Lemma. Let $W$ be a finite-dimensional complex inner product space and let $A,B\geq 0$ be positive semidefinite operators on $W$. For all unit vectors $w\in W$, it holds that
  $$ 
\langle w,(A+B)w\rangle \geq \lambda (1-\cos\theta),\tag{2}
$$ 
  where $\theta$ is the angle between $\ker A$ and $\ker B$, and $\lambda$ is the quantity defined above. 

How do we minimize the left-hand side of (2)? To make this as small as possible from a geometric-heuristic argument, we should choose $w$ to be as close to the kernels of both $A$ and $B$ as possible in order to maximize 
$$
\langle w,\, (\Pi_{\ker A}+\Pi_{\ker B})w\rangle.\tag{3}
$$
This will ensure that the overlap that $w$ has with $\Pi_{\operatorname{supp} A}$ and $\Pi_{\operatorname{supp} B}$ is as small as possible.
As earlier, suppose that $u\in\ker A$ and $v\in \ker B$ are the unit vectors satisfying 
$$
\langle u,v\rangle = \cos\theta,
$$
where $\theta=\theta(\ker A,\ker B)$ is the angle between the kernels of $A$ and $B$. To minimize the left side of (2) and maximize (3), we should choose $w = \alpha(u+v)$, where $\alpha = \frac{1}{\sqrt{2(1+\cos\theta)}}$. 
It clearly holds that $A\geq\lambda (I-\Pi_{\ker A})$ and $B\geq\lambda (I-\Pi_{\ker B})$, so for this choice of unit vector $w$ we have that
\begin{align*}
\langle w,(A+B)w\rangle 
&= \frac{1}{2(1+\cos\theta)}(\langle v, Av\rangle + \langle u, Bu\rangle)\\
&\geq \frac{\lambda}{2(1+\cos\theta)}\left(\langle v,(I-\Pi_{\ker A})v\rangle + \langle u,(I-\Pi_{\ker B})u\rangle\right)\\
& = \frac{\lambda}{2(1+\cos\theta)} (2-2\cos^2\theta)\\
& = \lambda\frac{1-\cos^2\theta}{1+\cos\theta}\\
& = \lambda(1-\cos\theta).
\end{align*}
Note. This isn't a geometric "proof", but rather a heuristic argument that helps explain Kitaev's lemma.
