Linear Algebra, 2020a

Topic: Orthonormal Matrices 321
(we assumed that t maps ⃗0 to itself)
( )
( )
( ) ( )
x
x
ax + cy a
dist( ,⃗e 1 )=dist(t( ),t(⃗e 1 )) = dist(
, )
y
y
bx + dy b
and
( )
( )
( ) ( )
x
x
ax + cy c
dist( ,⃗e 2 )=dist(t( ),t(⃗e 2 )) = 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 is a linear map plus a translation,
f(⃗v) =t(⃗v)+⃗v 0 for some constant vector ⃗v 0 and linear map t that is distancepreserving.
So in order to understand distance-preserving maps what remains is
to understand distance-preserving linear maps.
Not every linear map is distance-preserving. For example ⃗v ↦→ 2⃗v 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(⃗e 1 )‖ = ‖t(⃗e 2 )‖ = 1, and t(⃗e 1 ) is
orthogonal to t(⃗e 2 ). 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. To show the ‘if’ half we can check that the map preserves lengths
of vectors because then for all ⃗p and ⃗q the distance between the two is preserved
‖t(⃗p − ⃗q )‖ = ‖t(⃗p)−t(⃗q )‖ = ‖⃗p − ⃗q ‖. For that check let
( )
x
⃗v =
y
( )
a
t(⃗e 1 )=
b
( )
c
t(⃗e 2 )=
d
and with the ‘if’ assumptions that a 2 + b 2 = c 2 + d 2 = 1 and ac + bd = 0 we
have this.
‖t(⃗v )‖ 2 =(ax + cy) 2 +(bx + dy) 2
= a 2 x 2 + 2acxy + c 2 y 2 + b 2 x 2 + 2bdxy + d 2 y 2
= x 2 (a 2 + b 2 )+y 2 (c 2 + d 2 )+2xy(ac + bd)
= x 2 + y 2
= ‖⃗v ‖ 2
One thing that is neat about this characterization is that we can easily
recognize matrices that represent such a map with respect to the standard
bases: the columns are of length one and are mutually orthogonal. This is an
orthonormal matrix (or, more informally, orthogonal matrix since people