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!
For the first one, you are given that $(B-2I)(B-I)= (B-I)(B-2I) = 0$. To show that $W$ is the direct sum of two subspaces, you first show that the two subspaces intersect only at $0$, then show that $W$ is the sum of both subspaces.
For the sum part, let $w \in W$. Then, we know that $w = (B-I)w - (B - 2I)w$. Note that the first term belongs to $\ker(B-2I)$ and the second to $\ker(B-I)$. Thus, certainly $W = \ker(B-2I) + \ker(B-I)$.
To see trivial intersection, note that $Bw-2w = 0$ and $Bw - w = 0$ imply $w = 0$ by subtraction. Hence, the sum above is direct.
It would not be right to use dimensions here, since $\dim W$ and the ranks of $B-I,B-2I$ are not known.
An equivalent condition for diagonalizability of $F$ is that its minimal polynomial should have distinct roots. In this case, the minimal polynomial of $M$ over $\mathbb R$ certainly divides $P(x) = x^2 - 3x +2$, since $P(M) = 0$. Therefore, since $P(x)$ has distinct roots, we get that the minimal polynomial must also have distinct roots, and therefore that $M$ is diagonalizable. Note that this also implies the remark about algebraic and geometric multiplicity that you have made.