I have the next question:
Suppose that $1$ and $2$ are eigenvalues of a linear map $\phi:\mathbb{C}^3\rightarrow\mathbb{C}^3$. Moreover, suppose that $\phi$ is not diagonalizable. Let $p=p(x)$ denote the characteristic polynomial of $\phi$. What can be $p(x)$?
As far as I understand, if the matrix cannot be diagonalized, this means that there is no basis that consist of eigenvectors. But what does that tell me about the third eigenvalue? That it is $0$? Or that the third eigenvalue can be as equal to $1$ as equal to $2$? Give me some hints plz
The third eigen value is $1$ or $2$. Hence $p(x)=(x-1)(x-2)^{2}$ or $p(x)=(x-2)(x-1)^{2}$.