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Let $A$ be an $n$ by $n$ matrix over the field of all complex numbers and $\det(\lambda E - A)$ its characteristic equation. Suppose that $$ \det(\lambda E - A) = \lambda^n + c_1 \lambda^{n-1} + c_2 \lambda^{n-2} + \cdots + c_n. $$ Let $\lambda_1, \ldots, \lambda_n$ be the eigenvalues of $A$. Then we know that $\sum_{i=1}^{n} \lambda_i = c_1$ and $\lambda_1 \cdots \lambda_n = c_n$. Are there some formulas for $c_2, \ldots, c_{n-1}$? Thank you very much.

Edit: Then we know that $\sum_{i=1}^{n} \lambda_i = \det(A)$ and $\lambda_1 \cdots \lambda_n = Tr(A)$. Are there some formulas for $c_2, \ldots, c_{n-1}$ in terms of elements of $A$? Thank you very much.

LJR
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  • You might want to read a bit about elementary symmetric polynomials. I think that would solve your question – HSN May 06 '13 at 09:16
  • Actually, you have $(-\lambda_1)+\cdots+(-\lambda_n)=c_1$ and $(-\lambda_1)\cdots(-\lambda_n)=c_n$. And $\sum_{i=1}^n\lambda_i=\mathrm{Tr},A$ while $\prod_{i=1}^n\lambda_i=\det A$. – ccorn May 06 '13 at 11:13
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    In words, $(-1)^m,c_m$ is the sum of all $m\times m$ principal minors of $A$. E. g. for $n=3$, $m=2$, you have $c_2 = \left|\begin{array}{cc}a_{11}&a_{12}\a_{21}&a_{22}\end{array}\right| +\left|\begin{array}{cc}a_{11}&a_{13}\a_{31}&a_{33}\end{array}\right| +\left|\begin{array}{cc}a_{22}&a_{23}\a_{32}&a_{33}\end{array}\right|$. A proof strategy might employ the Faddeev-Leverrier method, but I will not try that here, therefore I post this as a comment only. – ccorn May 06 '13 at 12:50
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    See also this thread. – ccorn May 06 '13 at 19:49

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Since $\lambda_i$s are solutions of $\sum_{i=0}^{n} c_{n-i} \lambda^i = 0$, we can just use Vieta's formulas.

JiminP
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