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Problem

Let $\mathbb{v}$ and $\mathbb{w}$ be $\mathbb{Z}$-valued vectors with $n$ entries, and $X$ an $n\times n$ integer-valued matrix such that $$X\mathbb{v}=\mathbb{w},\hspace{10pt}X^T \mathbb{w}=\mathbb{v},$$ so that $$X^T X \mathbb{v}=\mathbb{v},\hspace{10pt} XX^T\mathbb{w}=\mathbb{w}.$$

I would like to prove $\mathbb{v}$ and $\mathbb{w}$ have the same entries up to permutation and signs, or find a counter-example.

Some ideas

If $XX^T=1$, I could prove that each line of $X\in O(n,\mathbb{Z})$ is composed of zeros and one $\pm 1$, since it is an integer vector with norm 1. This would imply what I want, but the assumption is too strong.

All I have is that $\mathbb{v}$ is an eigenvector of $X^T X$ with eigenvalue 1. I want to say something along the lines of the transformation $X$ being orthogonal in the subspace spanned by $\mathbb{v}$, but does that make sense? And does it imply anything for $\mathbb{w}$?

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    It doesn't quite make sense to say that $X$ is orthogonal on the subspace spanned by $\mathbb{v}$ because it is not even clear if $X$ acts on this one dimensional subspace (i.e., is $X\mathbb{v} $ in this subspace?). If $X$ does indeed act on this subspace, then $\mathbb{w}$ must be a scalar multiple of $\mathbb{v}$. – 2017-01-06
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    You have a good point. So this is not the way to go. I would welcome other ideas if anyone knows the solution. – 2017-01-06

1 Answers 1

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I have found a counter-example.

$$\mathbb{v}=\begin{pmatrix} 2\\ -1 \\-1\\0\\0\\0\end{pmatrix}, \hspace{10pt} \mathbb{w}=\begin{pmatrix} 1 \\ 1 \\ -1 \\ -1 \\ -1 \\ -1 \end{pmatrix},$$

related by

$$X=\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ \dots \end{pmatrix}$$