$T^2=I$ implies $T$ is diagonalizable Suppose $T:V\rightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.
First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?
 A: In this case you can split $V = V_1 \oplus V_{-1}$ where $T$ acts as either $1$ or $-1$.  Then take the direct sum over each subspace. $T = T_1 \oplus T_{-1}$

Since $T^2 v = v$ for all $v \in \mathbb{V}$ (an involution), we can decompose every vector as:
$$ v = \underbrace{\tfrac{1}{2}(v + Tv)}_{E_1} + 
\underbrace{\tfrac{1}{2}(v - Tv)}_{E_{-1}},$$
where $E_\lambda$ is an eigenspace with eigenvalue $\lambda$, so this eigenspace decomposition exists for all vector.
A: Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.
A: According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.
A: Your proposition is in general false unless you specify which elements the matrices are allowed to take.
The matrix $T = \left(\begin{array}{cc} 0&1\\1&0\end{array}\right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.
This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $\left(\begin{array}{cc} 1&1\\1&1\end{array}\right)$ we can perform both addition and multiplication - we have constructed new numbers! 
Example, add 2 and -1: Just ordinary matrix addition: $\left(\begin{array}{cc} 2&1\\1&2\end{array}\right)$, but after stripping the projection on $\left(\begin{array}{cc} 1&1\\1&1\end{array}\right)$, we are left with $\left(\begin{array}{cc} 1&0\\0&1\end{array}\right)$ which represents 1.
