In fact, we always have equality.

Suppose $v \in \text{Eig}(H_\mu, \mu)$.  Write $v = v_1 + v_2$ where 
$v_1 = H_1 v$, $v_2 = (I-H_1) v$ are orthogonal.
We have
$$ \mu v_2 = H_\mu v - \mu v_1 = \mu H_1 v + (1-\mu) H_2 v - \mu v_1 =  (1-\mu) H_2 v$$
i.e. $$H_2 v = \frac{\mu}{1-\mu} v_2$$
Now since $H_2$ is idempotent, $H_2 v_2 = v_2$, so that 
$$H_2 v_1 = H_2 v - v_2 = \dfrac{2\mu - 1}{1-\mu} v_2 $$
But by symmetry,
$$0 = v_1^T H_2 v_2 = v_2^T H_2 v_1 = \dfrac{2\mu - 1}{1-\mu} v_2^T v_2$$
and since $\mu \ne 1/2$, $v_2 = 0$, and $v = v_1 \in \text{img}(H_1) \cap \text{ker}(H_2)$.