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](https://math.stackexchange.com/questions/1672040/eigenspace-of-convex-combination-of-two-idempotent-matrices).



Eigenspace of convex combination of two idempotent matrices
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 $\text{ker}(H\_3)$.


