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Math 3191Applied Linear Algebra
Lecture 14: Coordinate Representations
Stephen Billups
University of Colorado at Denver
Math 3191Applied Linear Algebra – p.1/22
Tonight
Finish 4.3
Section 4.4: Coordinate Representations
Math 3191Applied Linear Algebra – p.2/22
Bases for Col A
EXAMPLE: Find a basis for Col A, where
A = [a1 a2 a3 a4] =
1 2 0 4
2 4 −1 3
3 6 2 22
4 8 0 16
.
Solution: Row reduce:
[a1 a2 a3 a4] ∼ · · · ∼
1 2 0 4
0 0 1 5
0 0 0 0
0 0 0 0
= [b1 b2 b3 b4]
Math 3191Applied Linear Algebra – p.3/22
Note thatb2 = b1 and a2 = a1
b4 = 4b1 + 5b3 and a4 = 4a1 + 5a3
b1 and b3 are not multiples of each other
a1 and a3 are not multiples of each other
Elementary row operations on a matrix do not affectthe linear dependence relations among the columnsof the matrix.
Therefore Span{a1, a2, a3, a4} =Span{a1, a3} and {a1, a3}is a basis for Col A.
Math 3191Applied Linear Algebra – p.4/22
THEOREM 6
The pivot columns of a matrix A form a basis for Col A.
EXAMPLE: Let v1 =
1
2
−3
,v2 =
−2
−4
6
,v3 =
3
6
9
.
Find a basis for Span{v1,v2,v3} .
Math 3191Applied Linear Algebra – p.5/22
Solution:
Let A =
1 −2 3
2 −4 6
−3 6 9
and note that
Col A = Span{v1,v2,v3}.
By row reduction, A ∼
1 −2 0
0 0 1
0 0 0
. Therefore a basis
for Span{v1,v2,v3} is
,
.
Math 3191Applied Linear Algebra – p.6/22
Review:
1. To find a basis for Nul A, use elementary row operationsto transform [A 0] to an equivalent reduced row echelonform [B 0]. Use the reduced row echelon form to findparametric form of the general solution to Ax = 0. Thevectors found in this parametric form of the generalsolution form a basis for Nul A.
2. A basis for Col A is formed from the pivot columns of A.Warning: Use the pivot columns of A, not the pivotcolumns of B, where B is in reduced echelon formand is row equivalent to A.
Math 3191Applied Linear Algebra – p.7/22
Section 4.4 Coordinate Systems
In general, people are more comfortable working with thevector space Rn and its subspaces than with other types ofvectors spaces and subspaces. The goal here is to imposecoordinate systems on vector spaces, even if they are not inRn.
THEOREM 7 The Unique Representation Theorem
Let β = {b1, . . . ,bn} be a basis for a vector space V . Thenfor each x in V , there exists a unique set of scalarsc1, . . . , cn such that
x = c1b1 + · · · + cnbn.
Math 3191Applied Linear Algebra – p.8/22
DEFINITION
Suppose β = {b1, . . . ,bn} is a basis for a vector space V
and x is in V . The coordinates of x relative to thebasis β (or the β− coordinates of x) are the weightsc1, . . . , cn such that x = c1b1 + · · · + cnbn.
In this case, the vector in Rn
[x]β =
c1
...cn
is called the coordinate vector of x (relative to β), or theβ− coordinate vector of x.
Math 3191Applied Linear Algebra – p.9/22
EXAMPLE
Let β = {b1,b2} where b1 =
2
4
3
1
3
5 and b2 =
2
4
0
1
3
5 and let E = {e1, e2} where
e1 =
2
4
1
0
3
5 and e2 =
2
4
0
1
3
5.
Solution: If [x]β =
2
4
2
3
3
5, then
x =____
2
4
3
1
3
5 + ____
2
4
0
1
3
5 =
2
4
3
5 .
If [x]E =
2
4
6
5
3
5, then x =____
2
4
1
0
3
5 + ____
2
4
0
1
3
5 =
2
4
3
5 .
Math 3191Applied Linear Algebra – p.10/22
Graphical Representation
1 2 3 4 5 6 7x1
1
2
3
4
5
6
7x2
Standard Graph Paper β−Graph Paper
Math 3191Applied Linear Algebra – p.11/22
From the last example,
2
4
6
5
3
5 =
2
4
3 0
1 1
3
5
2
4
2
3
3
5 .
Pβ = [b1 b2 · · · bn] and [x]β =
2
6
6
6
6
6
6
4
c1
c2
...
cn
3
7
7
7
7
7
7
5
Then
x =Pβ [x]β .
Math 3191Applied Linear Algebra – p.12/22
We call Pβ the change-of-coordinates matrix from β tothe standard basis in Rn.
Key Fact: Pβ is invertible. Why?
Thus, [x]β = P−1
β x
and therefore P−1
β is a change-of-coordinates matrix fromthe standard basis in Rn to the basis β.
Math 3191Applied Linear Algebra – p.13/22
EXAMPLE:
Let b1 =
2
4
3
1
3
5, b2 =
2
4
0
1
3
5 , β = {b1,b2} and x =
2
4
6
8
3
5.
(a) Find the change-of-coordinates matrix Pβ from β to the standard basis in R2 andchange-of-coordinates matrix P−1
βfrom the standard basis in R2 to β.
Solution Pβ = [b1 b2] =
2
4
3
5 and so P−1
β=
2
4
3 0
1 1
3
5
−1
=
2
4
1
30
− 1
31
3
5 .
(b) If x =
2
4
6
8
3
5, then use P−1
βto find [x]β =
2
4
2
6
3
5.
Solution: [x]β = P−1
βx =
2
4
1
30
− 1
31
3
5
2
4
6
8
3
5 =
2
4
3
5
Math 3191Applied Linear Algebra – p.14/22
Coordinate mappings allow us to introduce coordinatesystems for unfamiliar vector spaces.
Standard basis for P2 : {p1,p2,p3} ={
1, t, t2}
Math 3191Applied Linear Algebra – p.15/22
Polynomials in P2 behave like vectors in R3. Sincea + bt + ct2 = ____p1 + ____p2 + ____p3,
[
a + bt + ct2]
β=
a
b
c
We say that the vector space R3 is isomorphic to P2.
Math 3191Applied Linear Algebra – p.16/22
Parallel Worlds of R3 and P2
Vector Space R3 Vector Space P2
Vector Form:
2
6
6
4
a
b
c
3
7
7
5
Vector Form: a + bt + bt2
Vector Addition Example Vector Addition Example2
6
6
4
−1
2
−3
3
7
7
5
+
2
6
6
4
2
3
5
3
7
7
5
=
2
6
6
4
1
5
2
3
7
7
5
`
−1 + 2t − 3t2´
+`
2 + 3t + 5t2´
= 1 + 5t + 2t2
Math 3191Applied Linear Algebra – p.17/22
Informally, we say that vector space V is isomorphic toW if every vector space calculation in V is accuratelyreproduced in W , and vice versa.More precisely, vector spaces V and W are isomorphicif there is a one-to-one linear transformation from V ontoW .
Assume β is a basis set for vector space V . Exercise 25(page 254) shows that a set {u1,u2, . . . ,up} in V is
linearly independent if and only if{
[u1]β , [u2]β , . . . , [up]β
}
is linearly independent in Rn.
Math 3191Applied Linear Algebra – p.18/22
EXAMPLE
Use coordinate vectors to determine if {p1,p2,p3} is a linearly independent set, wherep1 = 1 − t, p2 = 2 − t + t2, and p3 = 2t + 3t2.
Solution: The standard basis set for P2 is β =˘
1, t, t2¯
. So
[p1]β =
2
6
6
4
3
7
7
5
, [p2]β =
2
6
6
4
3
7
7
5
, [p3]β =
2
6
6
4
3
7
7
5
Then2
6
6
4
1 2 0
−1 −1 2
0 1 3
3
7
7
5
∼ · · · ∼
2
6
6
4
1 2 0
0 1 2
0 0 1
3
7
7
5
By the IMT,n
[p1]β , [p2]β , [p3]β
o
is linearly ____________________ and therefore
{p1,p2,p3} is linearly ____________________. Math 3191Applied Linear Algebra – p.19/22
EXAMPLE
Coordinate vectors also allow us to associate vector spaceswith subspaces of other vectors spaces.
Let β = {b1,b2} where b1 =
2
6
6
4
3
3
1
3
7
7
5
and b2 =
2
6
6
4
0
1
3
3
7
7
5
and let H =span{b1,b2}. Find
[x]β , if x =
2
6
6
4
9
13
15
3
7
7
5
.
Math 3191Applied Linear Algebra – p.20/22
Solution
(a) Find c1 and c2 such that
c1
3
3
1
+ c2
0
1
3
=
9
13
15
Corresponding augmented matrix:
3 0 9
3 1 13
1 3 15
v
1 0 3
0 1 4
0 0 0
Therefore c1 = ____ and c2 = _____ and so [x]β =
[ ]
.
Math 3191Applied Linear Algebra – p.21/22
x1x2
x3
9
13
15
in R3 is associated
with the vector
[
3
4
]
in R2.
H is isomorphic to R2.
Math 3191Applied Linear Algebra – p.22/22
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