Page 114 - 35Linear Algebra

114 Linear Transformations

6.2 Linear Functions on Hyperplanes

It is not always so easy to write a linear operator as a matrix. Generally,
this will amount to solving a linear systems problem. Examining a linear
function whose domain is a hyperplane is instructive.

Example 71 Let
     
1 0
 

1
V = c 1   + c 2   c 1 , c 2 ∈ R
1

0 1
 
3
and consider L : V → R be a linear function that obeys
       
1 0 0 0
L   =   , L   =   .
1
1
1
1
0 0 1 0
By linearity this specifies the action of L on any vector from V as
      
1 0 0
1
1
1
L c 1   + c 2   = (c 1 + c 2 )   .

0 1 0
The domain of L is a plane and its range is the line through the origin in the x 2
direction.
It is not clear how to formulate L as a matrix; since
       
c 1 0 0 0 c 1 0
1
L c 1 + c 2  =  1 0 1   c 1 + c 2  = (c 1 + c 2 )   ,

c 2 0 0 0 c 2 0
or

       
c 1 0 0 0 c 1 0
1
L c 1 + c 2  =  0 1 0   c 1 + c 2  = (c 1 + c 2 )   ,

c 2 0 0 0 c 2 0
you might suspect that L is equivalent to one of these 3 × 3 matrices. It is not. By
3
the natural domain convention, all 3 × 3 matrices have R as their domain, and the
domain of L is smaller than that. When we do realize this L as a matrix it will be as a
3×2 matrix. We can tell because the domain of L is 2 dimensional and the codomain
is 3 dimensional. (You probably already know that the plane has dimension 2, and a

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