After applying a sequence of involutory real matrices to a vector, is the norm of this vector bounded from below? For $n,N \in \mathbb{N}$, let $A_1, \ldots, A_n$ be a finite sequence of involutory $(N \times N)$-matrices over $\mathbb{R}$, i.e. $A = A^{-1}$.
We know, that the eigenvalues of any involutory matrix lie in the set $\{-1,+1\}$. Further, each involution is diagonalizable, i.e. there are no generalized eigenvectors.
Assume that each matrix $A_i$, $i \in \{1, \ldots, n\}$, does not have all eigenvalues equal to $-1$ or $+1$ and let $i_k \in \{1, \ldots ,n\}$, for every $k \in \mathbb{N}$.
I conjecture that, for every $v \in \mathbb{R}^{N} \setminus \{\vec{0}\}$
\begin{align}
\lim_{k \to \infty} \|A_{i_k}A_{i_{k-1}} \ldots A_{i_1}v\|
\end{align}
is bounded away from zero.
Here is my (geometrical) try: For every $v \in \mathbb{R}^{N} \setminus \{\vec{0}\}$, there exist coordinates $\alpha_{1,{i_1}}, \ldots, \alpha_{N,{i_1}}$, such that
\begin{align}
v = \alpha_{1,{i_1}}v_{1,{i_1}} + \ldots + \alpha_{N,{i_1}}v_{N,{i_1}},
\end{align}
where $v_{1,{i_1}}, \ldots, v_{N,{i_1}}$ are the eigenvectors corresponding to the matrix $A_{i_1}$. When applying $A_{i_1}$ to $v$, we see that the eigenvectors corresponding to the eigenvalue $1$ stay put and the rest changes orientation. This means, $v$ is reflected on the eigenspace corresponding to eigenvalue $1$ and therefore does not decrease in length. Finish by induction.
Any thoughts?

Counterexample:
$$\frac{1}{\sqrt5}\begin{pmatrix}1&2\\2&-1\end{pmatrix} \begin{pmatrix}1&2\\0&-1\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\frac{1}{\sqrt5}\begin{pmatrix}2\\-1\end{pmatrix}$$ So each iteration of this couple divides the vector by $\sqrt5$, so converging to 0. The error in OP's argument is to assume the eigenvectors are orthogonal.
