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 $v = v_1 \in \text{img}(H_1) \cap \text{ker}(H_2)$.