Let $A$ and $B$ be two $4\times 4$ matrices. Using Newton's identities, one can prove that if
$$\det(A) = \det(B)\quad \text{and}\quad \mathrm{tr}(A^i) = \mathrm{tr}(B^i)$$ for $i=1,2,3$, then $A$ and $B$ have the same characteristic polynomial, thus the same eigenvalues.


I'm interested in pairs of matrices $A$ and $B$ that satisfy all those equations except the last one, i.e.
$$\det(A)=\det(B)$$
$$\mathrm{tr}(A)=\mathrm{tr}(B)$$
$$\mathrm{tr}(A^2)=\mathrm{tr}(B^2)$$
but $\mathrm{tr}(A^3) \neq \mathrm{tr}(B^3)$.


Does anyone know how to generate such matrices? Have they ever been studied? A reference would be nice.



On matrices that almost have the same eigenvalues
Such matrices will have a characteristic polynomial $z^4+a\_3z^3+a\_2z^2+a\_1z+a\_0$ with the same $a\_3$, $a\_2$, $a\_0$ but distinct $a\_1$. You can generate a plenty of diagonal such matrices by picking roots of such two polynomials.
 I cannot vouch that they were not studied but I am pretty certain that nothing groundbreaking came out of such studies.


