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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.

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    $\begingroup$ 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 :) $\endgroup$
    – t3suji
    Commented Jun 17, 2010 at 16:27
  • $\begingroup$ 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. $\endgroup$
    – Q.Q.J.
    Commented Jun 17, 2010 at 20:14
  • $\begingroup$ Perhaps it would be an interesting question to ask that all the eigenvalues of both matrices be integers? $\endgroup$
    – Gerry Myerson
    Commented Nov 4, 2020 at 22:15

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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.

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  • $\begingroup$ Thanks, that answers my question as stated. Though, I was also hoping for references that give properties of such matrices. Do you know any? $\endgroup$
    – Malik Younsi
    Commented Jun 17, 2010 at 16:58
  • $\begingroup$ Maybe you could be more specific in terms of what kind of properties you're looking for, or why you're interested in these matrices in the first place. The (very restrictive!) properties which define these matrices surely generate more identities than one could possibly know what to do with. $\endgroup$
    – Cam McLeman
    Commented Jun 17, 2010 at 17:06
  • $\begingroup$ The question is rather basic to be discussed specifically in a paper. You can refer to any book that does Jordan forms ad elementary symmetric functions, e.g., Artin's Algebra. $\endgroup$
    – Bugs Bunny
    Commented Jun 17, 2010 at 21:18

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