# On matrices that almost have the same eigenvalues

Let $A$ and $B$ be two $4$ by $4$ matrices. Using Newton's identites, one can prove that if $$det(A) = det(B)$$ and $$tr(A^i) = tr(B^i)$$ 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. $$det(A)=det(B)$$ $$tr(A)=tr(B)$$ $$tr(A^2)=tr(B^2)$$ but $tr(A^3) \neq tr(B^3)$.

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

Thank you, Malik

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Isn't that just saying that the characteristic poly's of A and B differ only in the linear term? That gives an easy way to generate such matrices :) – t3suji Jun 17 '10 at 16:27
Yes I was making some additional assumptions unthinkingly t3suji, thanks. Maybe I'll have another go if Malik tells us what kind of properties he/she is interested in. – Q.Q.J. Jun 17 '10 at 20:14

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.