About eigenvalues and complex matrix If $A$ is a square complex matrix with $n$ rows, prove that the constant term of the characteristic polynomial is equal to $(-1)^ndet(A)$ and that the coefficient of degree $n-1$ is equal to $-Tr(A)$
Thank you!
 A: Hint (for that $\det$-thing): for any polynomial $f$ of degree $n$ it's additive constant equals $f(0)$. Now plug in $\lambda=0$ in $\det(A-\lambda I)$.  
A: Recall that the characteristic polynomial of $A$ is $\det(zI_n-A)=(z-\lambda _1)\cdot \ldots \cdot(z-\lambda _n)$, for certain scalars $\lambda _1, \ldots ,\lambda _n$.
You'd like to show that $\det(A)=\lambda _1\cdot\ldots \cdot\lambda _n$.
Hint$_\bf1$: What can you pick to replace $z$ to help you with this?

Let $A=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}$.
Hint$_\bf 2$: Prove on the one hand that $$\det(zI_n-A)=z^n+b_{n-1}z^{n-1}+\ldots+b_1z+b_0,$$ for certain $b_{n-1}, \ldots ,b_0\in \mathbb C$, where $b_{n-1}=a_{11}+\ldots +a_{nn}$ and on the other hand that
$$(z-\lambda _1)\cdot \ldots \cdot(z-\lambda _n)=z^n+(\lambda_1+\ldots +\lambda _n)z^{n-1}+\ldots +(\lambda _1\cdot \ldots \cdot\lambda _n).$$
(Use Vieta's formulas).
