Linear motion in the Euclidean plane?

Question

This is what is said in my book on linear algebra, can you please give an explanation to this? I don't quite understand the notations that are used.

The set $\Bbb R^2$ can be viewed as the Euclidean plane. In this context, linear functions of the form $f :\Bbb R^2\rightarrow \Bbb R$ or$f :\Bbb R^2\rightarrow \Bbb R^2$ can be interpreted geometrically as “motions” in the plane and are called linear transformations.

I get it that a complex plane would have the set $i\times\Bbb R$ and the same reasoning would give that $\Bbb R^2$ is a plane in the Euclidean plane. And $f$ is obviously the function so my guess would be that $f :\Bbb R^2\rightarrow \Bbb R^2$ means that the function $f$ has a domain and co-domain that can reach from negative infinity to positive infinity (real numbers). Am I wrong? Or does it say that $f$ has been transformed from one Euclidean plane to another? So one frame of reference to another? So much confusion... Please someone sort this out. Thank you!

Or... Is it the fact that we have made another form of transformation using a vector? What I mean is, if we have a point Q if we apply the vector v on the point Q the point Q would then be transformed into another point in the same Euclidean plane?

Answer 1

An example of linear transformation with domain in $\mathbb R^2$ and image in $\mathbb R^2$ is a change of coordinates: for example, take a standard system of coordinates $(x,y)$ to a polar system of coordinates $(r,\theta)$. If you want to think in a physics point of view, think as two reference frames, one in motion relative to the other , and make a transformation (galilean or lorentzian, it doesn't matter in this case). There's two ways of making this transformation, which are equivalent:

1) Take a vector into another with a linear transformation, example: $${T(x,y) = (y,x)}$$ a rotation of the VECTOR $(x,y)$.

2) Take the COORDINATES of a vector in a system of coordinates to other coordinates in a linear way: $${v'_i = \sum_{k=1}^{2}a_{ik}v_k, i =1,2}$$ where the primed components are components of the SAME VECTOR in a different coordinate system, and the unprimed are the coordinates in the anterior coordinate system.

Answer 2

Your last guess might be the correct idea.$\def\rr{\mathbb{R}}$ "$f : \rr^2 \to \rr^2$" means that $f$ is a function from $\rr^2$ to $\rr^2$, equivalently given any vector $v \in \rr^2$, we have $f(v) \in \rr^2$. So if you treat $\rr^2$ as a plane, $f$ takes an input point in the plane and produces an output point in the plane. A translation of the plane is an example of such an $f$. Rotations too and reflections and scalings. It turns out that rotations and reflections are linear transformations, meaning that $f(u+v) = f(u) + f(v)$ for any $u,v \in \rr^2$ and $f(cv) = c f(v)$ for any $v \in \rr^2$ and (scalar) $c \in \rr$. There are other linear transformations too, and the most convenient representation of them is the matrix form. More specifically, given any linear transformation $f$ on $\rr^2$, there is a matrix $A$ such that $f(v) = Av$ for any $v \in \rr^2$. Furthermore, any $2 \times 2$ matrix $A$ gives a linear transformation $(v \mapsto Av)$ on $\rr^2$.

Answer 3

I don't know what you mean by "a complex plane would have the set $i\times \mathbb R$ and the same reasoning would give that $\mathbb R^2$ is a plane in the Euclidean plane." In what sense does a complex plane "have" a set? If you simply mean that we can think of the point $(a, b)$ in $\mathbb R^2$ as representing the complex number $a+ ib$, so that we have a "regular" real axis and the imaginary axis is "$i$" times all of the real numbers, yes, that is the idea

"And $f$ is obviously the function so my guess would be that $f:\mathbb R^2\rightarrow \mathbb R^2$ means that the function f has a domain and co-domain that can reach from negative infinity to positive infinity (real numbers)." Well, no, the domain and co-domain are sets of pairs of numbers "from negative infinity to positive infinity (real numbers)"

Answer 4

So, to build upon the answer by user21820, your book on linear algebra seems to state the situation the wrong way around, since scaling is not a geometric motion which preserves distance. The case of a linear transformation mapping into a lesser dimension is the extra case of adding a geometric projection (which is also a linear transformation). All linear transformations in 3 dimensional space can be represented geometrically by a rotation followed by a projection, a (optional) mirroring, a positive scaling, and a final rotation.