Page 246 - Matrix Analysis & Applied Linear Algebra

4.7 Linear Transformations 241
 
8
7
v =   ,
4
determine the coordinates of v with respect to the basis
      
1 1 1
 
1
B = u 1 =   , u 2 =   , u 3 =   .
2
2
1 2 3
 
Solution: The object is to find the three unknowns α 1 ,α 2 , and α 3 such that
α 1 u 1 + α 2 u 2 + α 3 u 3 = v. This is simply a 3 × 3 system of linear equations
         
111 α 1 8 α 1 9
122 = 7 =⇒ [v] B = = 2 .
   α 2     α 2   
123 α 3 4 α 3 −3
The general rule for making a change of coordinates is given on p. 252.

Linear transformations possess coordinates in the same way vectors do be-
cause linear transformations from U to V also form a vector space.

Space of Linear Transformations

• For each pair of vector spaces U and V over F, the set L(U, V)of
all linear transformations from U to V is a vector space over F.

• Let B = {u 1 , u 2 ,..., u n } and B = {v 1 , v 2 ,..., v m } be bases for U
and V, respectively, and let B ji be the linear transformation from
T
U into V defined by B ji (u)= ξ j v i , where (ξ 1 ,ξ 2 ,...,ξ n ) =[u] B .
That is, pick off the j th coordinate of u, and attach it to v i .
i=1...m
B L = {B ji } is a basis for L(U, V).
j=1...n
dim L(U, V) = (dim U) (dim V) .

Proof. L(U, V) is a vector space because the defining properties on p. 160 are
satisfied—details are omitted. Prove B L is a basis by demonstrating that it is a
linearly independent spanning set for L(U, V). To establish linear independence,

suppose η ji B ji = 0 for scalars η ji , and observe that for each u k ∈B,
j,i
m

v i if j = k
B ji (u k )= =⇒ 0= η ji B ji (u k )= η ji B ji (u k )= η ki v i .
0 if j = k
j,i j,i i=1
For each k, the independence of B implies that η ki = 0 for each i, and thus

B L is linearly independent. To see that B L spans L(U, V), let T ∈L(U, V),