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The characteristic equation is shown below

$p\left(\lambda\right) = \det\left(A-\lambda\mathbf{I}\right) = \left|\begin{array}{cccc} a_{11}-\lambda & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22}-\lambda & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}-\lambda \end{array}\right|$

$= \left(-1\right)^n \left[\lambda^n + c_1 \lambda^{n-1} + c_2 \lambda^{n-2} + \cdots + c_{n-1} \lambda + c_n\right]$

and it is identified the coefficient of $λ^n$ to be $(-1)^n$. but I not sure how we can prove this equal to $1$ (monic)

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  • $\begingroup$ The coefficient is sometimes -1, it depends on whether $n$ is even or odd (specifically when $n$ is even then the coefficient of $\lambda^n$ is 1, and when $n$ is odd the coefficient is -1). $\endgroup$ Sep 7 '16 at 7:47
  • $\begingroup$ The way you show the definition of the char. polynomial, the leading coefficient is always $\;(-1)^n\;$ , not $\;1\;$ . That's the main reason why I define the char. polynomial as $\;\det(\lambda I-A)\;$, and then yes: we always get this way a monic polynomial. $\endgroup$
    – DonAntonio
    Sep 7 '16 at 7:53
  • $\begingroup$ The proof relies on your definition of determinant: which one do you use? $\endgroup$
    – Crostul
    Sep 7 '16 at 7:56
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Expand the determinant in minors using Laplace's formula. You then see there is only one way of obtaining a term with $n$ powers of $\lambda$, namely multiplying the $n$ different $-\lambda$ together. You get $$(-\lambda)^n=(-1)^n \lambda^n.$$

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Using the definition of the characteristic polynomial that you have given the coefficient will be $(-1)^n$ not $1$. Inorder to always obtain a monic polynomial you can use the definition $$p(\lambda) = det(\lambda I-A)$$ That way the roots of the polynomials are the same (i.e. the eigenvalues can still be found correctly) but other properties will differ (e.g. it isn't necessarily monic).

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