About eigenvalue and a matrix Let $A$ be a $3\times3$  not-invertible matrix, and $\lambda=2$ is eigenvalue of $A$.
Then what can I say about $A$ and its eigenvalues?
$\lambda=0$ is also an eigenvalue of $A$ since its not invertible, but what about its polynomial?
Is it $\lambda^2(\lambda-2)$ or is it $\lambda(\lambda-2)^2$ ?
Thanks!

In order to do a complete analysis, you should study the three possible characteristic polynomials of $A$.
As you wrote in the question, since $A$ is not invertible, we have $0\in\Sigma(A)\implies \text {Ker}(F)\ne\{\underline 0\}$, where $F$ is the linear map associated to the matrix $A$.
You can easily observe this beacuse if $0\in\Sigma(A)\implies\exists\underline v\ne\underline 0:A\underline v=0\implies \underline v\in\text {Ker}( F)$, so the map is not injective.
The characteristic polynomial can be
$$\chi(t)=\begin{cases}-t(t-2)(t-k)&k\ne0,2\\t^2(t-2)\\-t(t-2)^2 \end{cases}.$$
