"If two matrices have the same trace and determinant, do they have the same characteristic polynomial?"

This is more of a number theory problem if we use the eigenvalue definition for trace and determinant and I have never studied any number theory before. If I gave you n numbers and told you the sum equals p and product equals q, and product equals p, and gave you another set of n numbers also with the same sum and product, are these sets the same set of numbers?

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    $\begingroup$ if the matrices are size 2 by 2, the characteristic polynomial really is $x^2 - Tx + D$ where $T$ is the trace and $D$ is the determinant. For 3 by 3 or bigger, the coefficient of $x$ is not determined by those $\endgroup$ – Will Jagy Mar 27 at 2:13
  • $\begingroup$ Concise perfect answer @WillJagy. $\endgroup$ – A rural reader Mar 27 at 2:28

Hint. Consider the $n\times n$ zero matrix and any $n\times n$ singular matrix whose trace is zero.


Every non-constant monic polynomial is the characteristic polynomial of its companion matrix.

It follows that when $n\ge3$, there always exist two $n\times n$ matrices with the same traces and determinants but different characteristic polynomials: just pick two monic degree-$n$ polynomials with equal coefficients of $x^{n-1}$ and equal constant terms, but different coefficients of $x^k$ for some $0<k<n-1$. Then their companion matrices will have equal traces and equal determinants, but different characteristic polynomials by construction.

When $n=2$, the characteristic polynomial of a matrix $A$ is $x^2-\operatorname{tr}(A)x+\det(A)$. Hence it is uniquely determined by the trace and determinant. The same conclusion holds when $n=1$, provided that the given trace is equal to the given determinant.


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