# 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?

• You can save the result, if you assume all $A_i$ to be symmetric. Jun 18, 2020 at 0:18

$$\frac{1}{\sqrt5}\begin{pmatrix}1&2\\2&-1\end{pmatrix} \begin{pmatrix}1&2\\0&-1\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\frac{1}{\sqrt5}\begin{pmatrix}2\\-1\end{pmatrix}$$ 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.
• I don't quite see where I used orthogonality, so let me rephrase my question with your counterexample: We have $A_2A_1v = \frac{1}{5}v$. Let $E_i$ be the matrix with columns of eigenvectors of the matrix $A_i$ and $[v]_{E_i}$ coordinates of $v$ wrt the corresponding eigenvector basis. I can rewrite $A_2A_1v = A_2E_2E_2^{-1}A_1E_1[v]_{E_1}$. Did you mean in your response that $E_2^{-1} = E_1^{-1}$ in order to obtain a lower bound? I don't quite see the orthogonal part. PS: I think $\sqrt{5}$ should be $5$ in your example otherwise the first matrix is not an involution. Jun 19, 2020 at 21:32
• The error is "v is reflected on the eigenspace corresponding to eigenvalue $1$ [should be $-1$] and therefore does not decrease in length". If you look at the counterexample, the vector does decrease in length after the reflection $A_1$. Btw it should be $\sqrt5$ not $5$; only after squaring $A_1^2$, should it become $1/5$. Jun 20, 2020 at 5:10
• Correction: ...only after squaring $A_2^2$... Jun 20, 2020 at 5:17