Linear Algebra

Topic: Orthonormal Matrices 287
That will follow immediately from this statement: a map t that is distancepreserving
and sends �0 to itself is linear. To prove this statement, let
� �
� �
a
c
t(�e1) = t(�e2) =
b
d
for some a, b, c, d ∈ R. Then to show that t is linear, it suffices to 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 any point in the plane can
be described by giving its distance from those three. So any point �v in the
domain is determined by its distance from the three fixed points �0, �e1, and�e2.
Similarly, any point t(�v) in the codomain is determined by its distance from the
three fixed points t(�0), t(�e1), and t(�e2) (these three are not collinear because, as
mentioned above, collinearity is invariant and �0, �e1, and�e2 are not collinear).
In fact, because t is distance-preserving, we can say more: for the point �v in the
plane that is determined by being the distance d0 from �0, the distance d1 from
�e1, and the distance d2 from �e2, its image t(�v) must be the unique point in the
codomain that is determined by being d0 from t(�0), d1 from t(�e1), and d2 from
t(�e2). Because of the uniqueness, checking that the action in (∗) works in the
d0, d1, andd2cases � �
� �
� �
x
x
ax + cy
dist( ,�0) = dist(t( ),t(�0)) = dist( ,�0)
y
y
bx + dy
(t is assumed to send �0 toitself)
� �
� �
� � � �
x
x
ax + cy a
dist( ,�e1) = dist(t( ),t(�e1)) = dist( , )
y
y
bx + dy b
and
� �
� �
� � � �
x
x
ax + cy c
dist( ,�e2) = dist(t( ),t(�e2)) = dist( , )
y
y
bx + dy d
suffices to show that (∗) describes t. Those checks are routine.
Thus, any distance-preserving f : R 2 → R 2 can be written f(�v) =t(�v)+�v0
for some constant vector �v0 and linear map t that is distance-preserving.
Not every linear map is distance-preserving, for example, �v ↦→ 2�v does not
preserve distances. But there is a neat characterization: a linear transformation
t of the plane is distance-preserving if and only if both �t(�e1)� = �t(�e2)� =1and
t(�e1) is orthogonal to t(�e2). The ‘only if’ half of that statement is easy—because
t is distance-preserving it must preserve the lengths of vectors, and because t
is distance-preserving the Pythagorean theorem shows that it must preserve
orthogonality. For the ‘if’ half, it suffices to check that the map preserves