Let $\mathbb{Z}_p$ be the ring of $p$-adic integers. Let $\mathrm{GL}_n(\mathbb{Z}_p)$ be the group of $n \times n$ invertible matrices with entries in $\mathbb{Z}_p$. Consider any matrix $M \in \mathrm{GL}_n(\mathbb{Z}_p)$ of order $2$, i.e. satisfying the polynomial $X^2 -1 \in \mathbb{Z}_p[X]$ with $M \neq 1$.
My ultimate hope is to show that such a matrix $M$ must be conjugate to a matrix of the form $$ D_{(m)} := \begin{pmatrix} -I_{(m)} & 0 \\ 0 & I_{(m-n)} \end{pmatrix}, $$ for some $1 \leq m \leq n$ in $\mathrm{GL}_n(\mathbb{Z}_p)$, where $I_{(k)}$ is the $k \times k$ identity matrix.
But I don't know how to show this. Could someone provide a proof of the above claim, or provide a counterexample?
My thoughts: If we turn to $\mathrm{GL}_n(\mathbb{Q}_p)$, since the minimal polynomial of such a matrix $M$ (can only be $X^2-1$ or $X+1$ in $\mathbb{Q}_p[X]$) has no multiple roots, the matrix $M$ is (conjugation) diagonalizable. Hence $M$ indeed conjugates to some $D_{(m)}$.
But when we are in $\mathrm{GL}_n(\mathbb{Z}_p)$, is such a matrix always (conjugation) diagonalizable? I have no idea on this.
General thoughts: So it seems the main obstruction is that I'm lacking a theory of doing linear algebra over any DVR, or even PID $R$. It seems that we cannot dream of it since when we regard our $R$-vector space $V$ as a $R[X]$-module with $X$ acting by the matrix $M$, $R[X]$ cannot be a PID, so we cannot use the structure theorem of finitely generated modules over PIDs. That is quite disappointing.
If my dream is not realistic, I'm hoping to see that every such matrices $M$ is conjugate in $\mathrm{GL}_n(\mathbb{Z}_p)$ to a matrix of the form $$ U_{(m)} := \begin{pmatrix} -I_{(m)} & \ast \\ 0 & I_{(m-n)} \end{pmatrix}, $$ where $\ast$ may not be zero.
Thank you all for answering and commenting!
If $p \ne 2$ then simply write
$$A = \{ (I+M) v,v\in \Bbb{Z}_p^n\}, \qquad B=\{ (I-M) v,v\in \Bbb{Z}_p^n\}$$ $$v=\frac12 (I+M) v +\frac12 (I-M) v\in A \oplus B$$ $A$ and $B$ are finitely generated thus free $\Bbb{Z}_p$-submodules of $\Bbb{Z}_p^n$ and $\Bbb{Z}_p^n=A\oplus B$.
$A$ is the $+1$ eigenspace and $B$ is the $-1$ eigenspace.
Taking a basis of $A$ and $B$ and putting the vectors in the columns of a matrix $P\in GL_n(\Bbb{Z}_p)$ gives a diagonalization $$M=P \pmatrix{I & \\ & -I} P^{-1}$$
If $p=2$ then $$\pmatrix{&1\\1&}\in GL_2(\Bbb{Z}_2)$$ is not diagonalizable over $GL_2(\Bbb{Z}_2)$, the eigenspaces are $\Bbb{Z}_2\pmatrix{1\\1},\Bbb{Z}_2\pmatrix{1\\-1}$ and their direct sum isn't the whole of $\Bbb{Z}_2^2$.