Let $H_1,H_2\in\mathbb{Q}^{n\times n}$ be idempotent and symmetric matrices. For any $0<\mu<\frac{1}{2}$, consider the matrix
$$H_\mu:=\mu H_1+(1-\mu)H_2.$$
I'm looking for a description of $\text{Eig}(H_\mu,\mu)$.
Clearly, $\text{img}(H_1)\cap\ker(H_2)\subseteq\text{Eig}(H_\mu,\mu)$, but under which conditions do we have equality? Is there a description of the "missing piece"
$$\text{Eig}(H_\mu,\mu)/\text{img}(H_1)\cap\ker(H_2)?$$
See also my question here.
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 $v = v_1 \in \text{img}(H_1) \cap \text{ker}(H_2)$.
EDIT: For larger convex combinations, an analogous statement is not true. For example, consider $H_\mu = \dfrac{1}{6} H_1 + \mu_2 H_2 + \left(\dfrac{5}{6} - \mu_2\right) H_3$ for the symmetric idempotents $$ \eqalign{H_1 &= \pmatrix{1 & 0\cr 0 & 0\cr},\cr H_2 &= \pmatrix{9/34 & -15/34\cr -15/34 & 25/34\cr}, \cr H_3 &= \pmatrix{25/26 & -5/26\cr -5/26 & 1/26\cr}} $$ The condition for $H_\mu$ to have eigenvalue $1/6$ is $8712 \mu_2^2 - 9108 \mu_2 + 2125 = 0$, which has a root near $\mu_2 = .3514749310$. Then you can check that for this $\mu_2$, $H_\mu$ has an eigenvector for eigenvalue $ 1/6$ which is not in $\text{img}(H_1)$ or $\text{ker}(H_2)$ or $\text{ker}(H_3)$.
Let me rephrase the answer in different terms.
Since $H_i$ are idempotent and self-adjoint, they are orthogonal projectors onto some subspaces $W_i$. Set $U=W_1\cap W_2$, and let $U_i$ be the orthogonal complement of $U$ in $W_i$. Finally, let $U^\perp$ be the orthogonal complment of $W_1+W_2$ in the whole space $V$. Then $V=U\oplus U_1\oplus U_2\oplus U^\perp$ (but $U_i$ are not necessarily orthogonal to each other). Notice here that $H_i(U_{3-i})\subseteq U_i$ by orthogoality.
Now, each vector $v$ is uniquely expanded as $v=u+u_1+u_2+u^\perp$ with each summand lying in the corresponding space. If $v$ is an eigenvector of $H_\mu$ with the eigenvalue $\mu$, then $$ \mu u+\mu u_1+\mu u_2+\mu u^\perp =H_\mu v =u+\mu u_1+(1-\mu)H_2(u_1)+(1-\mu)u_2+\mu H_1(u_2) =u+\mu(u_1+H_1(u_2))+(1-\mu)(u_2+H_2(u_1))+0, $$ the terms in the right-hand part also lie in the corresponding subspaces. Hence $u=u^\perp=0$, $H_1(u_2)=0$ (so $u_2\perp U_1$), and $(2\mu-1)u_2=(1-\mu)H_2(u_1)$. The last equality cannot hold unless $u_2=0$ or $\mu=1/2$, since $u_2\perp U_1$. So $v=u_1\in \mathop{\rm Im} H_1$ and $H_2(u_1)=0$ (thus $u_1\in\mathop{\rm Ker} H_2$), as required.
Perhaps, this language is better for generalizations?
