Compute the equivalence classes

Question

Define an equivalence relation on $\mathbb{R}^2$ by $\textbf{x}\sim\textbf{y}$ iff $\exists A\in GL_2(\mathbb{R})$ such that $A\mathbf{x}=\mathbf{y}$. Compute the equivalence classes of this equivalence relation.

My attempt:

Let $\mathbf{x}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$.

$A\mathbf{x}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ $\forall A\in GL_2(\mathbb{R})$

So, it seems that the zero vector resides alone in its equivalence class.

My hunch is that all the other (nonzero) vectors reside in the other equivalence class, making a total of 2 equivalence classes. But I don't know how to prove this as there doesn't seem to be any obvious way to solve for the matrix $A$ in the equation $A\mathbf{x}=\mathbf{y}$.

Can someone please tell me how to proceed?

Answer 1

Consider $x=\begin{pmatrix}1\\1\end{pmatrix}$.

Consider $A=\begin{pmatrix}\lambda & 0\\0&\mu\end{pmatrix}$, where $\lambda$, $\mu$ are nonzero so that $A$ is invertible and thus in $GL_2(\mathbb{R})$.

Then $Ax=\begin{pmatrix}\lambda\\\mu\end{pmatrix}$.

So all vectors $\begin{pmatrix}\lambda\\\mu\end{pmatrix}$, with $\lambda,\mu$ both nonzero are in the same class as $\begin{pmatrix}1\\1\end{pmatrix}$.

The final question is how about those vectors with one component nonzero? We can see that they are also in the same equivalence class:

$\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}c\\0\end{pmatrix}=\begin{pmatrix}c\\c\end{pmatrix}\sim\begin{pmatrix}1\\1\end{pmatrix}$

$\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}0\\c\end{pmatrix}=\begin{pmatrix}c\\c\end{pmatrix}\sim\begin{pmatrix}1\\1\end{pmatrix}$

Answer 2

Thanks, yoyostein!

I think I have a geometric solution.

Let $\mathbf{x}$ and $\mathbf{y}$ be any 2 nonzero points in $\mathbb{R}^2$. Let $\mathbf{x}=(r_1, \theta_1)$ and $\mathbf{y}=(r_2, \theta_2)$ in polar coordinates.

To transform point $\mathbf{x}$ to point $\mathbf{y}$ by matrix multiplication, we must rotate $\mathbf{x}$ counterclockwise by $\theta=\theta_2-\theta_1$ and expand or contract it by a factor of $\displaystyle\frac{r_2}{r_1}$.

To rotate, we multiply by the matrix $B=\begin{bmatrix} \cos(\theta_2-\theta_1) & -\sin(\theta_2-\theta_1) \\ \sin(\theta_2-\theta_1) & \cos(\theta_2-\theta_1) \end{bmatrix}$

To expand or contract, we multiply by the matrix $C=\begin{bmatrix} \displaystyle\frac{r_2}{r_1} & 0 \\ 0 & \displaystyle\frac{r_2}{r_1} \end{bmatrix}$

Since both these matrices are invertible, their product $A=BC$ is also invertible. So, $A\mathbf{x}=\mathbf{y}$ where $A$ is invertible. So, any two nonzero points reside in the same equivalence class.

Answer 3

Every nonzero vector is part of a basis.

Given $x,y \in \mathbb R^2$ nonzero vectors, let $X=\{x,x'\}$ be an ordered basis containing $x$ and $Y=\{y,y'\}$ be an ordered basis containing $y$. Let $A=YX^{-1}$. Then $Ax=y$.