Assuming that they are both Hermetian, positive definite and have the same full rank. (To show the converse that if two matrices have the same eigenvalues, they must have the same determinant is easy.) However, I want to know if there is an good way to show that if they two matrices have the same determinant and same trace it does not imply that they will have the same (real) eigenvalues. I tried to think of an quick example but came up only with one where the determinant is zero. Any ideas?
Edit: I know it's true for $2 \times 2$ matrices which is easy to prove by the characteristics polynomial. However, it should not be for $3 \times 3$ matrices? What about a $3\times 3$ proof by contradiction? Note that the determinant should not be zero, as both matrices are positive definite by assumption, which excludes eigenvalues of zero. Also, we state nothing about the order of the eigenvalues.
Best Answer
Just try any random example. For instance, $A=\operatorname{diag}(1,3,5)$ and $B=\operatorname{diag}(4,\frac{5+\sqrt{10}}2,\frac{5-\sqrt{10}}2)$.