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Let $S$ be the set of $3 \times 3$ matrices $A$ satisfying the following conditions:

  1. All minors are $>0$ (i.e., $A$ is a strictly totally positive matrix);
  2. all principal minors are $>1$, except for the $3\times 3$ minor $\mathrm{det}(A)$, which is assumed to be $1$.

By general theory of total positivity, the eigenvalues of $A$ are simple and positive.

Question: Given positive numbers $\lambda_1>\lambda_2>\lambda_3$ whose product is $1$, does it exist a matrix $A$ in the set $S$ whose eigenvalues are exactly the $\lambda_i$'s?

Some crude computer simulations indicate that the answer is no, but this may be a mistake.

Remark: It is known that strictly totally positive matrices can have any simple positive spectrum (see Barrett and Johnson, Possible spectra of totally positive matrices. Linear Algebra Appl. 62 (1984), 231–233).

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