Let
- ${A_j} \in {\mathbb{C}^{n \times n}},\quad(j = 0,1,2,\dots,m)$
- ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + \dots+{A_1}\lambda + {A_0}$ is a matrix polynomial, and $\lambda $ is a complex variable.
Lemma
For any given analytic curve $\xi :\mathbb{R} \to \mathbb{C}$, the eigenvalues of $P(\lambda)P(\lambda)^*$ can be arranged in such way that, for all $j$, ${S_j}(\xi (t))$ are real analytic functions of $t\in \mathbb{R}$.
Furthermore, if $S_j(\lambda)$ be eigen value of $P(\lambda)P(\lambda)^*$ and $s_n(\lambda) = \min_j(S_j(\lambda))^{1/2}$, is a non-zero simple singular value of $P(\lambda)$ and $u_\lambda, v_\lambda$¸ are associated left and right singular vectors, respectively, then (writing $\lambda= x + iy$), $s_n(.)$ is a real analytic function in a neighbourhood of $\lambda$ and, gradient of ${S_n}(x + iy)$ equal,
$$\nabla {S_n}(x + iy) = \left( \Re \left(u_\lambda^*\frac{{\partial (x + iy)}}{{\partial (x)}}{v_\lambda } \right), \Re \left(u_\lambda ^*\frac{{\partial (x + iy)}}{{\partial (y)}}{v_\lambda } \right)\right)$$
Note: $s_n(\lambda)$ means $s_n(P(\lambda))$.
Question
How can we prove by this lemma, that:
If ${S_j}(\lambda ) = {S_k}(\lambda )$ for $j\ne k$ and for all $\lambda$ in a non-empty open set $U$, then $U=\mathbb{C}$ ?