Eigenvalues problem 1 
For an $n\times n$ matrix $A,$ if $$\det(A-\lambda I)=(\lambda_1 -\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda)=0,$$ 
  holds then we have
  $$
\begin {align*}
(\lambda_1 I-A)(\lambda_2 I-A)\cdots (\lambda_n I -A)=0
\end{align*}
$$

Since $Tr(A) = \sum_{k=1}^n \lambda_k$ and $\det(A) = \prod _{k=1}^n \lambda_k,$
I checked the case at n=2.
However, in general for $n,$ I don't know how to prove that.
Any help is appreciated!!
Thank you.
 A: We can construct a set of basis in the $n$-dimensional vector space using the eigenvectors of matrix $A$.  So any vector can be written as $$x = \sum_{i=1}^n x_iv_i,$$ where ${v_i}$ are the eigenvectors of $A$. Note that $$(\lambda_jI-A)v_j = 0 \quad \forall j,$$ then for any vector $x$, $$\begin{split}
  & (\lambda_1 I-A)(\lambda_2 I-A)\cdots (\lambda_n I -A)x \\
= & (\lambda_1 I-A)(\lambda_2 I-A)\cdots (\lambda_n I -A)\sum_{i=1}^nx_iv_i \\
= & (\lambda_1 I-A)(\lambda_2 I-A)\cdots (\lambda_{n-1} I -A)\sum_{i=1}^{n-1}x_iv_i \\
= & \ldots \\
= & 0.
\end{split}$$
This means matrix $(\lambda_1 I-A)(\lambda_2 I-A)\cdots (\lambda_n I -A)$ equals to 0.

For defective matrix which its eigenvectors doesn't span the whole space, we need to use the generalized eigenvectors. For eigenvalue $\lambda$ of matrix $A$ with $m$ multiplicity, it has $m$ linearly independent general eigenvectors $u_k$ satisfying $$(\lambda I-A)^mu_k = 0.$$
A complete basis of $n$ dimensional space can be formed with the set of regular eigenvectors and generalized eigenvectors. Also note that $(\lambda_iI-A)(\lambda_jI-A) = (\lambda_jI-A)(\lambda_iI-A)$, we can always arrange so the factors $(\lambda I-A)$ corresponding to multiple roots stay together.  The logic of the original proof still works.
