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?

  • 1
    $\begingroup$ You can save the result, if you assume all $A_i$ to be symmetric. $\endgroup$
    – Philipp123
    Jun 18, 2020 at 0:18

1 Answer 1



$$\frac{1}{\sqrt5}\begin{pmatrix}1&2\\2&-1\end{pmatrix} \begin{pmatrix}1&2\\0&-1\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\frac{1}{\sqrt5}\begin{pmatrix}2\\-1\end{pmatrix}$$ 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.

  • $\begingroup$ 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. $\endgroup$ Jun 19, 2020 at 21:32
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    $\begingroup$ 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 $1/5$. $\endgroup$ Jun 20, 2020 at 5:10
  • $\begingroup$ Correction: ...only after squaring $A_2^2$... $\endgroup$ Jun 20, 2020 at 5:17
  • $\begingroup$ Excuse the delay, I got it, thanks for your answer. $\endgroup$ Jul 13, 2020 at 20:33

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