Linear Transformations   

An important class of transformations are the linear transformations, which are of the form

T( u,v) = á au+bv,cu+dv ñ

Linear transformations are so named because they map lines through the origin in the uv-plane to lines through the origin in the xy-plane.

If points ( u,v) in the plane are associated with column matrices, [ u,v] ^t, then the linear transformation T(u,v) = á au+bv,cu+dv ñ can be written in matrix form as

T æ ç è u v ö ÷ ø = é ê ë a b c d ù ú û é ê ë u v ù ú û

The matrix of coefficients a,b,c,d is called the matrix of the transformation.       

EXAMPLE 5    Find the image of the unit square under the linear transformation

T æ ç è u v ö ÷ ø = é ê ë 2 1 1 1 ù ú û é ê ë u v ù ú û

Solution: Since linear transformations map straight lines to straight lines, we need only find the image of the 4 vertices of the unit square. To begin with, the point ( 0,0) is mapped to (0,0) . Associating the point ( 1,0) to the column vector [ 1,0] ^t yields

T æ ç è 1 0 ö ÷ ø = é ê ë 2 1 1 1 ù ú û é ê ë 1 0 ù ú û = é ê ë 2 1 ù ú û

Thus, the point ( 1,0) is mapped to the point (3,-2) . Likewise, associating ( 0,1) with [0,1] ^t leads to

T æ ç è 0 1 ö ÷ ø = é ê ë 2 1 1 1 ù ú û é ê ë 0 1 ù ú û = é ê ë 1 1 ù ú û

and associating ( 1,1) with [ 1,1] ^t leads to

T æ ç è 1 1 ö ÷ ø = é ê ë 2 1 1 1 ù ú û é ê ë 1 1 ù ú û = é ê ë 3 2 ù ú û

That is, ( 0,1) is mapped to ( 1,1) and (1,1) is mapped to ( 3,2) . Thus, the unit square in the uv-plane is mapped to the parallelogram in the xy-plane with vertices ( 0,0) , ( 2,1) , ( 1,1) , and (3,2) .

       

Among the most important linear transformations are rotations about the origin through an angle q, which are given by

T( u,v) = á u cos( q) - v sin(q) ,u sin( q) + v cos( q) ñ

(1)

The matrix of the rotation through an angle q is given by

R( q) = é ê ë cos( q) -sin( q) sin( q) cos( q) ù ú û

when positive angles are those measured counterclockwise (see the exercises).           

EXAMPLE 6    Rotate the triangle with vertices ( 0,0), ( 2,0) , and ( 0,2) through an angle q = p/3 about the origin.       

Solution: To begin with, the matrix of the rotation is

R( q) = é ê ê ê ê ê ë cos æ è  p 3 ö ø -sin æ è  p 3 ö ø sin æ è  p 3 ö ø cos æ è  p 3 ö ø ù ú ú ú ú ú û = é ê ë 1/2 -Ö3/2 Ö3/2 1/2 ù ú û

so that the resulting linear transformation is given by

T æ ç è u v ö ÷ ø = é ê ë 1/2 -Ö3/2 Ö3/2 1/2 ù ú û é ê ë u v ù ú û

The point ( 0,0) is mapped to ( 0,0) . The point ( 2,0) is associated with [ 2,0] ^t, so that

T æ ç è 2 0 ö ÷ ø = é ê ë 1/2 -Ö3/2 Ö3/2 1/2 ù ú û é ê ë 2 0 ù ú û = é ê ë 1 Ö3 ù ú û

That is, ( 2,0) is mapped to ( 1,Ö3) . Similarly, it can be shown that ( 0,2) is mapped to ( -Ö3,1) :