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Suppose $A$ and $B$ belong to a kind of special hermitian matrices, which have the following properties:

  1. $A$ and $B$ contain only one negative eigenvalue.
  2. the negative eigenvalue and the second-largest positive eigenvalue are opposite to each other.
  3. $\operatorname{trace}(A) = \operatorname{trace}(B)= 1$.

Let $$ C=\alpha A+(1-\alpha)B,\quad \alpha \in [0,1]. $$ Then, we can find that the absolute value of the minimum eigenvalue of $C$ is always less than the second-largest positive eigenvalue of $C$ (just like an upper bound).

Can this observation be proved?

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    $\begingroup$ This does not make it clear how it fits into qa-quantum-algebra or quantum-groups. $\endgroup$
    – Simon Willerton
    Commented Nov 4, 2021 at 8:23
  • $\begingroup$ oh, my fault, thanks. $\endgroup$
    – Mecheal Baker
    Commented Nov 4, 2021 at 13:14
  • $\begingroup$ It doesn't look correct even for diagonal matrices. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 11, 2021 at 1:28

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