Prove that there is a basis of a lattice $\Lambda$ s.t. a reflection is of a certain form

Question

Consider $\Lambda$ a lattice in $\mathbb{R}^2$. Let $S \in O(\Lambda)$ be a reflection, i.e. $\det S = -1$. Set $S_{1}= \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$ and $S_{2}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$.

I now want to prove that there is a basis of $\Lambda$ such that the matrix of $S$ is $S_1$ or $S_2$.

Until now I was only able to prove that since $S, S_1, S_2$ all have the same spectrum and thus I can deduce that there is a Basis of $\mathbb{R}^2$ such that the condition is fulfilled.

Question: How can I now prove that this basis is also a basis of $\Lambda$?

Any help is very appreciated!

Definition of a lattice and its basis:

$\lambda_1 \mathbb{Z} \oplus \dots \oplus \lambda_n \mathbb{Z} = \{m_1 \lambda_1 + \dots m_n \lambda_n | m_i \in \mathbb{Z} \} \subset V$ is the lattice generated by the linearly independent vectors $\lambda_1, \dots, \lambda_n \subset V$. Those are the basis.

Let $ \Lambda = \lambda_1 \mathbb{Z} \oplus \dots \oplus \lambda_n \mathbb{Z}$ and $ \Lambda' = \lambda_1' \mathbb{Z} \oplus \dots \oplus \lambda_n' \mathbb{Z}$

$\Lambda = \Lambda'$ if and only if there is a $C \in GL(n, \mathbb{Z}$ such that $\lambda_i' = \sum\limits_{j=1}^n C_{ij} \lambda_j$ for all $n$.

Answer 1

Here is one way to do it (I will write $L$ instead of $\Lambda$):

We already know that $S$ has Eigenvalues $1,-1$ and (due to those being integral) there is a basis $\{l_1,l_2\}\subset L$ of $\mathbb{R}^2$ such that $Sl_1=l_1$ and $Sl_2=-l_2$. We will take $l_1$ and $l_2$ reduced, i.e. $\frac{m}{n} l_i \notin L$ for $n \nmid m $.

Now there are two possibilities:

$(l_1,l_2)$ is already a basis of $L$. Then $S$ takes diagonal form with respect to this basis and we are done.

Otherwise there is some $l \in L - \langle l_1,l_2 \rangle_\mathbb{Z}$. But certainly $l \in \langle l_1,l_2 \rangle_\mathbb{Q}$ so write $l=\frac{a_1}{b_1}l_1+\frac{a_2}{b_2}l_2$ (both fractions completly reduced).

Then $b_1l-a_1l_1=\frac{a_2b_1}{b_2} l_2$ is an element of $L$, hence $b_2 \mid b_1$ and by an analogous computation $b_1 \mid b_2$. So $b_1=b_2=:b$.

Now $l+Sl=\frac{2a_1}{b}l_1$ is in $L$ hence $b=2$.

So every element of $L$ which is not in the sublattice generated by $l_1$ and $l_2$ is of the form $\frac{1}{2}(a_1l_1+a_2l_2)$ with $a_1,a_2$ odd.

But in that case $\frac{1}{2}(l_1+l_2), \frac{1}{2}(l_1-l_2)$ is a basis of $L$ and $S$ takes the second form with respect to this basis.

Edit: I just realized one can completly skip the step where I concluded $b_1=b_2$ and just have a look at $l+Sl$ and $l-Sl$ to get to $b_1=b_2=2$.

Answer 2

The definition of a reflection is not $\det(S) = -1$. For example, $\pmatrix{2&0\\0&-1/2}$ is not a reflection.

What, then, is the definition of a reflection?

Once you have a proper definition, you'll be able to state that since the matrices $S,S_1,S_2$ have the same spectrum (which consists of $n$ distinct eigenvectors), they are similar, which means that the desired statement holds.