A First Course in Linear Algebra, 2017a

5.8. The Matrix of a Linear Transformation II 311
To show that T is one to one, let T (∑ n k=1 c k⃗v k )=⃗0. Then ∑ n k=1 c kT (⃗v k )=∑ n k=1 c k⃗w k =⃗0. It follows
that each c k = 0 because it is given that {⃗w 1 ,···,⃗w n } is linearly independent. Hence T (∑ n k=1 c k⃗v k )=⃗0
implies that ∑ n k=1 c k⃗v k =⃗0 andsoT is one to one.
To show that T is onto, let ⃗w be an arbitrary vector in R n . This vector can be written as ⃗w = ∑ n k=1 d k⃗w k =
∑ n k=1 d kT (⃗v k )=T (∑ n k=1 d k⃗v k ). Therefore, T is also onto.
♠
Consider now an important definition.
Definition 5.56: Coordinate Vector
Let B = {⃗v 1 ,⃗v 2 ,···,⃗v n } be a basis for R n and let ⃗x be an arbitrary vector in R n .Then⃗x is uniquely
represented as⃗x = a 1 ⃗v 1 + a 2 ⃗v 2 + ···+ a n ⃗v n for scalars a 1 ,···,a n .
The coordinate vector of⃗x with respect to the basis B, written C B (⃗x) or [⃗x] B ,isgivenby
⎡ ⎤
a 1
a 2
C B (⃗x)=C B (a 1 ⃗v 1 + a 2 ⃗v 2 + ···+ a n ⃗v n )= ⎢ . ⎥
⎣ .. ⎦
a n
Consider the following example.
Example 5.57: Coordinate Vector
{[ ] [ ]}
[
1 −1
Let B = , beabasisofR
0 1
2 and let ⃗x =
3
−1
]
beavectorinR 2 .FindC B (⃗x).
Solution. First, note the order of the basis is important so label the vectors in the basis B as
{[ ] [ ]}
1 −1
B = , = {⃗v
0 1
1 ,⃗v 2 }
Now we need to find a 1 ,a 2 such that⃗x = a 1 ⃗v 1 + a 2 ⃗v 2 ,thatis:
[ ] [ ] [ ]
3 1 −1
= a
−1 1 + a
0 2
1
Solving this system gives a 1 = 2,a 2 = −1. Therefore the coordinate vector of ⃗x with respect to the basis
B is
[ ] [ ]
a1 2
C B (⃗x)= =
a 2 −1
Given any basis B, one can easily verify that the coordinate function is actually an isomorphism.
♠