Let $B$$:$ $W$ $→$ $W$ be a linear map such that $B^2-3B+2I = 0$, where $I$ is the identity map on $W$.
a) Prove that $W$ $=$ $Ker(B - 2I)$ $⊕$ $Ker(B - I)$.
b) Let $M$ be an $n$ x $n$ matrix such that $M^2-3M+2In$ $=$ $0$, where $In$ is the $n$ x $n$ identity matrix. Is $M$ diagonalizable?
For a), I understand that $Im(B-2I)$ $⊆$ $Ker(T-I)$ and $Im(B-I)$ $⊆$ $Ker(B-2I)$. And then I thought that because the intersection of the two images would be zero as they defined two different spaces, the intersection of the kernels would also be $0$. As for dimension, I thought that each kernel would be dimension $1$ and summing them together would give dimension $2$, which I think is the dimension of $W$ as there are two variables in the initial equation.
For b), I factorized the polynomials to get $(M-2I)(M-I)$ $=$ $0$. I thought that this equation could be used as the determinant equation for eigenvalues as it equaled $0$ and the determinant of the zero matrix is $0$. So I got from this that the eigenvalues were $2$ and $1$ and therefore the matrix is diagonalizable as the algebraic and geometric multiplicity are both $1$ for each eigenvalue.
But obviously, I am quite unsure if my reasoning and computation of the dimension are correct for a), and I am not sure if treating the polynomial as the eigenvalue equation in b) is the right way to go either.
Any help would be highly appreciated!