Linear Algebra

290 Chapter Three. Maps Between Spacessuffices to show that (∗) describes t. Those checks are routine.Thus we can write any distance-preserving f: R 2 → R 2 as f(⃗v) = t(⃗v) + ⃗v 0for some constant vector ⃗v 0 and linear map t that is distance-preserving. Sowhat is left in order to understand distance-preserving maps is to understanddistance-preserving linear maps.Not every linear map is distance-preserving. For example ⃗v ↦→ 2⃗v does notpreserve distances.But there is a neat characterization: a linear transformation t of the planeis distance-preserving if and only if both ‖t(⃗e 1 )‖ = ‖t(⃗e 2 )‖ = 1, and t(⃗e 1 ) isorthogonal to t(⃗e 2 ). The ‘only if’ half of that statement is easy — because tis distance-preserving it must preserve the lengths of vectors and because tis distance-preserving the Pythagorean theorem shows that it must preserveorthogonality. To show the ‘if’ half we can check that the map preserves lengthsof 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( )x⃗v =y( )at(⃗e 1 ) =b( )ct(⃗e 2 ) =dand with the ‘if’ assumptions that a 2 + b 2 = c 2 + d 2 = 1 and ac + bd = 0 wehave 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 ‖ 2One thing that is neat about this characterization is that we can easilyrecognize matrices that represent such a map with respect to the standardbases: the columns are of length one and are mutually orthogonal. This is anorthonormal matrix or orthogonal matrix (people often use the second termto mean not just that the columns are orthogonal but also that they have lengthone).We can use this to understand the geometric actions of distance-preservingmaps. Because ‖t(⃗v )‖ = ‖⃗v ‖, the map t sends any ⃗v somewhere on the circleabout the origin that has radius equal to the length of ⃗v. In particular, ⃗e 1 and ⃗e 2map to the unit circle. What’s more, once we fix the unit vector ⃗e 1 as mappedto the vector with components a and b then there are only two places where ⃗e 2can go if its image is to be perpendicular to the first vector’s image: it can mapeither to one where ⃗e 2 maintains its position a quarter circle clockwise from ⃗e 1