Direct Sum and Diagonalization with a Linear Map

Question

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!

Answer 1

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.

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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.