Dominant eigenvector of convex combination of Hermitian matrices

Question

Given Hermitian matrices ${\bf A}, {\bf B}, {\bf C} \in \mathbb{C}^{N \times N}$, we have the following optimization problem in vector ${\bf x} \in \mathbb{C}^N$

$$\begin{array}{ll} \text{maximize} & {\bf x}^H {\bf A} {\bf x}\\ \text{subject to} & \|{\bf x}\|^2_2 \leq 1\end{array}$$

which has a solution given by (using KKT conditions)

$${\bf A} {\bf x} = \mu {\bf x}$$

where $\mu > 0$, which indicates that the maximizer is the eigenvector corresponding to the dominant eigenvalue.

Suppose now that ${\bf A} = (1-\alpha) {\bf B} + \alpha {\bf C}$, where $0 \leq \alpha \leq 1$. Then we have

$$((1-\alpha) {\bf B} + \alpha {\bf C}) {\bf x} = \mu {\bf x}$$

In this case, can anyone suggest now to proceed with the solution for any $0 \leq \alpha \leq 1$ in terms of the eigenvectors of ${\bf B}$ and ${\bf C}$?

I use KKT conditions to reach the conclusion that the optimal ${\bf x}$ is the dominant eigenvector. I am stuck when there is a convex combination of the Hermitian Matrices, i.e. $(1-a){\bf B} + a{\bf C}$. — Chandan Pradhan, Dec 06 '19 at 11:48
You don't need KKT to reach that conclusion. Take a look at this. — Rodrigo de Azevedo, Dec 06 '19 at 11:56
I am still not sure how to find a general solution for $(1-\alpha) {\bf B} + \alpha {\bf C} {\bf x} = \mu {\bf x}$ in terms of eigenvectors. Actually, my primary problem let me reach $(1-\alpha) {\bf B} + \alpha {\bf C} {\bf x} = \mu {\bf x}$ hence I am seeking solution in terms of eigenvector of $(1-\alpha) {\bf B} + \alpha {\bf C}$. — Chandan Pradhan, Dec 06 '19 at 12:34
The problem may be too hard. Take a look at this. The AMS article is here. — Rodrigo de Azevedo, Dec 06 '19 at 12:39
Somewhat related: https://math.stackexchange.com/q/3209720/339790 — Rodrigo de Azevedo, Dec 06 '19 at 12:44

Dominant eigenvector of convex combination of Hermitian matrices

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