$M^2=I_n$ implies $M$ diagonalizable

Question

Let $M$ be an $n \times n$ matrix such that $M^{2}=I_n$; does this imply that $M$ is similar to a diagonal matrix $D$? How can we prove this?

Answer 1

Theorem:

Let $T$ be a linear operator on the finite-dimensional vector space $V$. then $T$ is diagonalizable iff minimal polynomial is of the form $g(t)=\prod_{k=1}^{n}(t-\lambda_k), \lambda_j \ne\lambda_k$ for any distict index.

Solution to the Problem

Given that $M^2=I.$ Let $g(t)=t^2-1$. We know that $g(M)=0$. Hence it is the minimal polynomial. $g(t)=(t-1)(t+1)$. Hence, $M$ is diagonalizable.

Answer 2

HINT: Show that the eigenspaces $\mathbf E(1)$ and $\mathbf E(-1)$ span $\Bbb R^n$.

Answer 3

Yes it is diagonalisable. Let $W_1$ be the space of generalized eigenvalues for $1$, that is vectors such that $(M-I)^nv=0$ for some $n$. Now note that $(M+I)W_1=W_1$ since otherwise there is an element in $W_1$ with eigenvalue $-1$. This implies that $$(M-I)W_1=(M-I)(M+I)W_1=0$$ this all generalized eigenvectors are actual eigenvectors. This implies that algebraic and geometric multiplicities are the same. Hence diagonalizable.

Answer 4

If $v\in F^n$ (where $F$ is the field that you're working with), then$$v=\frac12(v+M.v)+\frac12(v-M.v).$$Since $M^2=\operatorname{Id}$, this expresses $v$ as the sum of a vector $v_1$ such that $M.v_1=v_1$ with a vector $v_{-1}$ such that $M.v_{-1}=-v_{-1}$. So, every vector is sum of eigenvectors and therefore $M$ is diagonalizable.