A First Course in Linear Algebra, 2017a

518 Vector Spaces
Definition 9.86: Coordinate Isomorphism
Let V be a vector space with dim(V) =n, letB = { ⃗ b 1 , ⃗ b 2 ,..., ⃗ b n } be a fixed basis of V, andlet
{⃗e 1 ,⃗e 2 ,...,⃗e n } denote the standard basis of R n . We define a transformation C B : V → R n by
⎡ ⎤
a 1
C B (a 1
⃗ b1 + a 2
⃗ b2 + ···+ a n
⃗ a 2
bn )=a 1 ⃗e 1 + a 2 ⃗e 2 + ···+ a n ⃗e n = ⎢ .
⎣ ..
⎥
⎦ .
a n
Then C B is a linear transformation such that C B ( ⃗ b i )=⃗e i , 1 ≤ i ≤ n.
C B is an isomorphism, called the coordinate isomorphism corresponding to B.
We continue with another related definition.
Definition 9.87: Coordinate Vector
Let V be a finite dimensional vector space with dim(V) =n, andletB = { ⃗ b 1 , ⃗ b 2 ,..., ⃗ b n } be an
ordered basis of V (meaning that the order that the vectors are listed is taken into account). The
coordinate vector of⃗v with respect to B is defined as C B (⃗v).
Consider the following example.
Example 9.88: Coordinate Vector
Let V = P 2 and⃗x = −x 2 − 2x + 4. FindC B (⃗x) for the following bases B:
1. B = { 1,x,x 2}
2. B = { x 2 ,x,1 }
3. B = { x + x 2 ,x,4 }
Solution.
1. First, note the order of the basis is important. Now we need to find a 1 ,a 2 ,a 3 such that ⃗x = a 1 (1)+
a 2 (x)+a 3 (x 2 ),thatis:
−x 2 − 2x + 4 = a 1 (1)+a 2 (x)+a 3 (x 2 )
Clearly the solution is
a 1 = 4
a 2 = −2
a 3 = −1
Therefore the coordinate vector is
⎡
C B (⃗x)= ⎣
4
−2
−1
⎤
⎦