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I am looking for a way to relate the terms of the characteristic polynomial of a $3 \times 3$ matrix to its eigenvalues.

The definition I start with (taken from Wolfram MathWorld) is

$\\P_{3}(A)=x^{3} + \mbox{Tr}(A)x^{2} + (\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2}))x^1 +(\mbox{Tr}^{3}(A) + 2 \mbox{Tr} (A^{3}) -3 \mbox{Tr} (A) \mbox{Tr}(A^{2}))x^0$

Which can also be written as

$\\P_{3}(A)=x^{3} + \mbox{Tr}(A)x^{2} + (\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2}))x^1 + \det(A)x^0$,

which again can be rewritten using the properties of eigenvalues as

$\\P_{3}(A)=x^{3} + x^{2} \sum_{i=1}^{i=3} \lambda_{i} + x^1(\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2})) + x^0 \prod_{i=1}^{i=3} \lambda_{i}$ .

I am wondering is there a way to relate the term in $x$ also to the eigenvalues in a similar way that can be done in with the terms in $x^{2}$ and $x^{0}$?

Thanks!

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Yes. We have

$$\operatorname{Tr}^2(A)-\operatorname{Tr}(A^2)=(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)=2(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_1\lambda_3)$$

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  • $\begingroup$ It's that simple?! Eugh!! Thanks!! :) $\endgroup$ – QuantumPenguin Jun 19 '16 at 10:23
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    $\begingroup$ @QuantumPenguin More generally, the coefficients of the characteristic polynomial are elementary symmetric functions of the eigenvalues given by Vieta’s formulas. $\endgroup$ – amd Jun 19 '16 at 16:23
  • $\begingroup$ Thanks @amd, that's actually really useful for what I am trying to achieve! $\endgroup$ – QuantumPenguin Jun 21 '16 at 20:28

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