I is well known that there is no explicit formula for the eigenvalues of a general matrix (see e.g. Wikipedia). This result is a consequence of (1) Abel's theorem, stating that there is no explicit formula for the roots of a general polynomial of degree >=5, and (2) the fact that for every polynomial $p$ there is a matrix (the companion matrix) which has characteristic polynomial $p$.

My question: Is it true that there is no explicit formula for the eigenvalues of a general Hermitian matrix? Or in other words: is there a way to find, for a given polynomial $p$, a Hermitian matrix which has characteristic polynomial $p$?

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    $\begingroup$ It is sufficient to take $p$ with non-solvable Galois group. Take your favorite non-solvable real quintic polynomial. Now the diagonal matrix with diagonal entries the real roots of this polynomial is hermitian. This question is not research level in its present form. E.g. you should precise what you mean by "explicit formula" and on which field you work, to avoid stupid trivial answers as the one I just gave. $\endgroup$ Mar 16, 2015 at 10:10
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    $\begingroup$ I suspect that the OP is more interested in matrices with coefficients in $\mathbb{Q}$ or $\mathbb{Z}$. Though it would certainly help to make this explicit in the question if that is what is asked for. $\endgroup$ Mar 16, 2015 at 10:34
  • $\begingroup$ I apologise for off-topic-ness. Also, I meant matrices with coefficients in R, but since the question is on hold there seems little point to add this information now. I'll try Math.StackExchange in the future. $\endgroup$ Mar 16, 2015 at 19:28

1 Answer 1


A real symmetric tridiagonal matrix with a given characteristic polynomial, G. Schmeisser (1993)

earlier work (behind a paywall):

Expressing a polynomial as the characteristics polynomial of a symmetric matrix, M. Fiedler (1990)

  • $\begingroup$ Thanks, I had somehow failed to find this via Google, but these references solve my problem. $\endgroup$ Mar 16, 2015 at 19:30

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