20211104 why is the trace of a matrix equal to the sum of eigenvalues, and why is the determinant of a matrix equal to the product of eigenvalues

Why is the trace of a matrix equal to the sum of eigenvalues

seek $n$ Order matrix $A={\left(a_{ij}\right)}_{n\times n}$ The eigenvalues of the
$$\det (\lambda I-A)=\left|\begin{array}{cccc}\lambda -a_{11}& -a_{12}& \dots & -a_{1n}\\ -a_{21}& \lambda -a_{22}& \dots & -a_{2n}\\ \dots & \dots & \dots & \dots \\ -a_{n1}& -a_{n2}& \dots & \lambda -a_{nn}\end{array}\right|$$

The characteristic polynomial can be obtained from the expansion rule of determinant
$$\begin{array}{rlr}\phi (\lambda )& =\det (\lambda I-A)& \\ & ={\lambda }^n-\left(a_{11}+a_{22}+\cdots +a_m\right){\lambda }^{n-1}+& \cdots +(-1{)}^n\det \mathbf{A}\end{array}$$
meanwhile ,$\det (\lambda I-A)$ Yes $\mathrm{n}$ A root , They are $\mathrm{n}$ Eigenvalues , in other words
$$\det (\lambda I-A)=\left(\lambda -{\lambda }_1\right)\left(\lambda -{\lambda }_2\right)\dots \left(\lambda -{\lambda }_n\right)$$

that ,
$${\lambda }_1{\lambda }_2\cdots {\lambda }_n=\det \mathbf{A}$$
$$\mathrm{tr}\mathbf{A}=\sum _{i=1}^n{a}_{iu}$$
Reference resources ：
[1] https://www.zhihu.com/question/267405336
[2] 《 Matrix theory 》 Cheng Yunpeng Zhang Kaiyuan