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Let $A,B$ be two real $2 \times 2$ matrices with identical singular values $0<\sigma_1<\sigma_2$ and with a positive determinant (which is $\sigma_1 \sigma_2$).

Is there a known characterization for when the singular values of $tA+(1-t)B$ are again $\sigma_1,\sigma_2$ for every $t \in (0,1)$?

In particular, does this condition force $A=B$?

Does the answer change if we require this only for some specific (single) value of $t \in (0,1)$?

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Summary: Having fixed singular values implies being on a sphere, and the Euclidean unit ball is a strictly convex set.

Indeed, after some further thought, it seems that even if the singular values of $tA+(1-t)B$ are $\sigma_1,\sigma_2$ even for a single $t \in (0,1)$, then $A=B$.

Indeed, suppose that $C:=tA+(1-t)B$ has singular values $\sigma_1,\sigma_2$. Then

$$|C|=|A|=|B|=\sqrt{\sigma_1^2+\sigma_2^2}=t|A|+(1-t)|B|.$$

Thus, witing $\tilde A=tA, \tilde B=(1-t)B$, we get

$$ |\tilde A+\tilde B|=|\tilde A|+|\tilde B|,$$

so $\tilde A=\lambda \tilde B$ for some positive $\lambda$. Since we assumed that $t \in (0,1)$, we get $A=rB $ for some positive $r$, which must then be $1$, since $|A|=|B|>0$.

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