1. Coordinate System
Hung-yi Lee

2. Outline Coordinate Systems
Each coordinate system is a “viewpoint” for vector representation.
The same vector is represented differently in different coordinate systems.
Different vectors can have the same representation in different coordinate systems.
Changing Coordinates
Reference: textbook Ch 4.4

3. Using Basis to represent Vector
Coordinate System
Using Basis to represent Vector

4. Vector
𝑏 1 = 1 1
𝑏 2 = −1 1
6𝑏 1
8 4
8 4
4𝑒 2
6 −2
𝑏 1
𝑒 2
𝑏 2
𝑒 1
8𝑒 1
−2𝑏 2
8 4 =8 𝑒 1 +4 𝑒 2
8 4 =6 𝑏 1 +(−2) 𝑏 2

5. Vector 6 −2 8 4 drink home 𝑏 1 𝑒 1 𝑒 2 𝑏 2 𝑒 1 = 1 0 𝑒 2 = 0 1
𝑒 1 = 1 0
𝑒 2 = 0 1
𝑏 1 = 1 1
𝑏 2 = −1 1

6. Vector
𝑒 1 = 1 0
𝑒 2 = 0 1
𝑏 1 =
𝑏 2 =
2 3
2 3
for left
𝑒 2
𝑏 2
𝑏 1
𝑒 1
2 𝑏 1 +3 𝑏 2 =

7. Vector 𝑒 1 = 1 0 𝑒 2 = 0 1 𝑏 1 = 2 0.5 𝑏 2 = 0 0.5 2 3 2 3 home home
𝑒 1 = 1 0
𝑒 2 = 0 1
𝑏 1 =
𝑏 2 =
2 3
2 3
home
home
for left
𝑒 2
𝑏 2
𝑏 1
𝑒 1
2 𝑏 1 +3 𝑏 2 =

8. Coordinate System
A vector set B can be considered as a coordinate system for Rn if:
1. The vector set B spans the Rn
2. The vector set B is independent
Every vector should have representation
Unique representation
B is a basis of Rn

9. Why Basis? Let vector set B= 𝑢 1 , 𝑢 2 ,⋯, 𝑢 𝑘 be independent.
Any vector v in Span B can be uniquely represented as a linear combination of the vectors in B.
That is, there are unique scalars 𝑎 1 , 𝑎 2 ,⋯, 𝑎 𝑘 such that 𝑣= 𝑎 1 𝑢 1 + 𝑎 2 𝑢 2 +⋯+ 𝑎 𝑘 𝑢 𝑘
Proof:
Unique?
𝑣= 𝑎 1 𝑢 1 + 𝑎 2 𝑢 2 +⋯+ 𝑎 𝑘 𝑢 𝑘
𝑣= 𝑏 1 𝑢 1 + 𝑏 2 𝑢 2 +⋯+ 𝑏 𝑘 𝑢 𝑘
𝑎 1 −𝑏 1 𝑢 𝑎 2 −𝑏 2 𝑢 2 +⋯+ 𝑎 𝑘 −𝑏 𝑘 𝑢 𝑘 =0
B is independent
a1  b1 = a2  b2 =  = ak  bk = 0

10. Coordinate System 𝑣 B = B is a coordinate system
Let vector set B= 𝑢 1 , 𝑢 2 ,⋯, 𝑢 𝑛 be a basis for a subspace Rn
For any v in Rn, there are unique scalars 𝑐 1 , 𝑐 2 ,⋯, 𝑐 𝑛 such that 𝑣= 𝑐 1 𝑢 1 + 𝑐 2 𝑢 2 +⋯+ 𝑐 𝑛 𝑢 𝑛
B is a coordinate system
(某種觀點)
根據某種觀點所看到的結果
B -coordinate vector of v:
𝑣 B =
(用 B 的觀點來看原來的 v)

11. Coordinate System B= 𝑢 1 , 𝑢 2 ,⋯, 𝑢 𝑛 𝑣 B E= 𝑒 1 , 𝑒 2 ,⋯, 𝑒 𝑛 vector
(standard vectors)
E is Cartesian coordinate system
(直角坐標系)
𝑣= 𝑣 E

12. Other System → Cartesian
B= , 1 −1 1 ,
𝑣 B = 3 6 −2
𝑣= −1 1 −
= 7 −7 5
C= , ,
𝑢 C = 3 6 −2
𝑢= − =

13. Other System → Cartesian
Let vector set B= 𝑢 1 , 𝑢 2 ,⋯, 𝑢 𝑛 be a basis for a subspace Rn
Matrix B= 𝑢 1 𝑢 2 ⋯ 𝑢 𝑛
𝑣 B = 𝑐 1 𝑐 2 ⋮ 𝑐 𝑛
Given 𝑣 B , how to find v?
𝑣= 𝑐 1 𝑢 1 + 𝑐 2 𝑢 2 +⋯+ 𝑐 𝑛 𝑢 𝑛
=𝐵 𝑣 B
(matrix-vector product)

14. Cartesian → Other System
𝑣= 1 −4 4
B= , 1 −1 1 ,
find [v]B
[v]B= 𝑐 1 𝑐 2 𝑐 3
𝐵= −
B is invertible (?)
independent
= −6 4 3
𝑣 B = 𝐵 −1 𝑣
𝐵 𝑣 B =𝑣

15. Cartesian ↔ Other System
Let B = {b1 , b2 , , bn}
𝑣 B = 𝐵 −1 𝑣 𝑣
𝑣 B
= 𝑐 1 𝑐 2 ⋮ 𝑐 𝑛
𝑣=𝐵 𝑣 B
= c1b1 + c2b2 +  + cnbn
Let B= 𝑏 1 , 𝑏 2 ,⋯, 𝑏 𝑛 is a basis of Rn. 𝑏 𝑖 B =?
𝑒 𝑖
(Standard vector)

16. Changing Coordinates

17. Equation of ellipse ? Rotate 45◦ 2 2 3 3 Draw by ?

18. Equation of ellipse B = { 2 2 2 2 , − 2 2 2 2 } 𝑏 1 𝑏 2
Use another coordinate system
B = { , − } 2 3
𝑏 2
𝑏 1 is
What is the equation of the ellipse in the new coordinate system?

19. Equation of ellipse
B = { , − }

20. Equation of hyperbola 𝑦′ 𝑦′ 𝑦 𝑥′ 30◦ 𝑥 𝑥′ − 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12
在這裡鍵入方程式。
− 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12
𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛?

21. Equation of hyperbola 𝐵= 𝑏 1 𝑏 2 𝑦′ 𝑏 1 = 3 2 1 2 𝑏 2 = − 1 2 3 2 𝑦 𝑥′
𝐵= 𝑏 1 𝑏 2 𝑦′
𝑏 1 =
𝑏 2 = − 𝑦 𝑥′
30◦
𝑣 B = 𝑥′ 𝑦′
𝑣= 𝑥 𝑦
𝑣= B 𝑣 B 𝑥
𝑥 𝑦 = − 𝑥′ 𝑦′
− 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12

22. Equation of hyperbola 𝐵= 𝑏 1 𝑏 2 − 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12 𝑏 1 = 3 2 1 2
𝐵= 𝑏 1 𝑏 2
− 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12
𝑏 1 =
𝑏 2 = −
𝑥= 𝑥 ′ − 1 2 𝑦 ′
𝑦= 1 2 𝑥 ′ 𝑦 ′
𝑣 B = 𝑥′ 𝑦′
𝑣= 𝑥 𝑦
𝑣= B 𝑣 B
𝑥 ′ 𝑦 ′ =3
𝑥 𝑦 = − 𝑥′ 𝑦′

23. Summary B= 𝑢 1 , 𝑢 2 ,⋯, 𝑢 𝑛 𝑣 B E= 𝑒 1 , 𝑒 2 ,⋯, 𝑒 𝑛 vector vector
(standard vectors)
𝑣 B = 𝐵 −1 𝑣 𝑣
𝑣 B
𝑣=𝐵 𝑣 B

24. Appendix

25. Linear Algebra Review– Section 4.4 in one page
Given a basis B = {b1, b2, …, bn} for the vector space Rn. R2

26. Vector The same vector Represented by purple arrows
Represented by red arrows