# When the singular values of a convex sum are preserved?

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 $$t \in (0,1)$$?

## 1 Answer

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 $$|A|=|B|>0$$.