I have the exercise below and I can't do the last item.
Remember that an operator $S$ is an involution if $S^2=\text{Id}$. Let $T:\mathbb R^n \rightarrow \mathbb R^n$ be a linear operator. Show that:
a) $T$ is an involution if and only if $\mathbb R^n=V_0\oplus V_1$, with $T_{\mid V_0} = \text{Id}$ and $T_{\mid V_1}=-\text{Id}$.
b) If $T$ is an involution then $T$ is diagonalizable.
c) If $T$ is an involution then $T$ is normal if and only if $V_0 \perp V_1$.
d) Let $T_1, \dots, T_k$ distinct involution in $\mathbb R^n$ such that $T_iT_j=T_jT_i$, for any $i, j \in \{1, \dots, k\}$. Show that $k\leq 2^n$.
I completed items a, b and c , but can't start doing the d item. Any suggestions?
From part b) we know, that involutions are diagonalizable. By assumption, they are commuting. So $T_1, ..., T_k$ are simultaneously diagonalizable! Therefore there exist a basis $v_1,...,v_n$ of $\mathbb{R}^n$ of eigenvectors of $T_1,...,T_k$, i.e. $T_iv_l = (-1)^{m(i,l)}v_l$, where $i = 1,...,k$ and $l = 1,...,n$ and $m(i,l) \in \{0,1\}$. Now, $T_i = T_j$ if and only if $m(i,\cdot) = m(j, \cdot)$. But $m(i,\cdot) \in \{0,1\}^{n}$ and $|\{0,1\}^{n}| = 2^n$, so there are at most $2^n$ distinct functions $m(i,\cdot)$ and therefore at most $2^n$ distinct involutions in $\mathbb{R}^n$.