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)
Best Answer
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 $$(-\lambda)^n=(-1)^n \lambda^n.$$