Linear Algebra, 2020a

320 Chapter Three. Maps Between Spaces
under these maps. (This forms part of Klein’s Erlanger Program, which proposes
the organizing principle that we can describe each kind of geometry — Euclidean,
projective, etc. — as the study of the properties that are invariant under some
group of transformations. The word ‘group’ here means more than just ‘collection’
but that lies outside of our scope.)
We can use linear algebra to characterize the distance-preserving maps of
the plane.
To begin, observe that there are distance-preserving transformations of the
plane that are not linear. The obvious example is this translation.
( )
x
y
↦→
( ) ( ) ( )
x 1 x + 1
+ =
y 0 y
However, this example turns out to be the only one, in that if f is distancepreserving
and sends ⃗0 to ⃗v 0 then the map ⃗v ↦→ f(⃗v)−⃗v 0 is linear. That will
follow immediately from this statement: a map t that is distance-preserving
and sends ⃗0 to itself is linear. To prove this equivalent statement, consider the
standard basis and suppose that
( )
a
t(⃗e 1 )=
b
( )
c
t(⃗e 2 )=
d
for some a, b, c, d ∈ R. To show that t is linear we can show that it can be
represented by a matrix, that is, that t acts in this way for all x, y ∈ R.
( ) ( )
x t ax + cy
⃗v = ↦−→
(∗)
y bx + dy
Recall that if we fix three non-collinear points then we can determine any point
by giving its distance from those three. So we can determine any point ⃗v in the
domain by its distance from ⃗0, ⃗e 1 , and ⃗e 2 . Similarly, we can determine any point
t(⃗v) in the codomain by its distance from the three fixed points t(⃗0), t(⃗e 1 ), and
t(⃗e 2 ) (these three are not collinear because, as mentioned above, collinearity is
invariant and ⃗0, ⃗e 1 , and ⃗e 2 are not collinear). Because t is distance-preserving
we can say more: for the point ⃗v in the plane that is determined by being the
distance d 0 from ⃗0, the distance d 1 from ⃗e 1 , and the distance d 2 from ⃗e 2 ,its
image t(⃗v) must be the unique point in the codomain that is determined by
being d 0 from t(⃗0), d 1 from t(⃗e 1 ), and d 2 from t(⃗e 2 ). Because of the uniqueness,
checking that the action in (∗) works in the d 0 , d 1 , and d 2 cases
( )
( )
( )
x
x
ax + cy
dist( ,⃗0) =dist(t( ),t(⃗0)) = dist(
,⃗0)
y
y
bx + dy