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This question already has an answer here:

Given a $3\times 3$ symmetric matrix

\begin{equation} M= \begin{pmatrix} A & B & C \\ B & D & E \\ C & E & F\\ \end{pmatrix}, \end{equation}

how do I find the eigenvalues? Brute-forcing it by looking at the roots of the characteristic polynomial seems overwhelming.

Are there any properties of symmetric matrices that might simplify the problem or am I overlooking something?

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marked as duplicate by Jennifer, Watson, egreg, Kushal Bhuyan, mickep Jun 13 '16 at 12:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Lordofdark very relevant, but it doesn't provide a definitive answer to the question. $\endgroup$ – Omnomnomnom Jun 13 '16 at 12:21
  • $\begingroup$ Hi. Yes, I read it before I posted, but like @Omnomnomnom stated, they do not provide a clear answer. $\endgroup$ – Henrymerrild Jun 13 '16 at 12:22
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Unfortunately, there is (to my knowledge) no pencil and paper method for computing exact eigenvalues that exploits the symmetry of a matrix.

It is notable, however, that the eigenvalues of a symmetric matrix will necessarily be real. So, at the very least, that's a fact you can use to check for mistakes.

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In the general$^*$ case there is no escape, you must solve a cubic equation. (I am not talking about particular cases such that the coefficients and roots are rational or solutions can be spotted by inspection.)

Because whatever method you will use will deliver those roots, and will constitute an equivalent of Cardano's formulas. In addition, as the three roots are real, you can be in the casus irreductibilis an need trigonometric functions.


$^*$Note that the matrix has $6$ independent coefficients. This is more than enough to correspond to a general cubic polynomial.

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