Coordinate System Hung-yi Lee

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

Using Basis to represent Vector Coordinate System Using Basis to represent Vector

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

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

Vector 𝑒 1 = 1 0 𝑒 2 = 0 1 𝑏 1 = 2 0.5 𝑏 2 = 0 0.5 2 3 2 3 4 2.5 for left 𝑒 2 𝑏 2 𝑏 1 𝑒 1 2 𝑏 1 +3 𝑏 2 = 4 2.5

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 = 2 0.5 𝑏 2 = 0 0.5 2 3 2 3 home home 4 2.5 for left 𝑒 2 𝑏 2 𝑏 1 𝑒 1 2 𝑏 1 +3 𝑏 2 = 4 2.5

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

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 1 + 𝑎 2 −𝑏 2 𝑢 2 +⋯+ 𝑎 𝑘 −𝑏 𝑘 𝑢 𝑘 =0 B is independent a1  b1 = a2  b2 =  = ak  bk = 0

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)

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

Other System → Cartesian B= 1 1 1 , 1 −1 1 , 1 2 2 𝑣 B = 3 6 −2 𝑣=3 1 1 1 +6 1 −1 1 −2 1 2 2 = 7 −7 5 C= 1 2 3 , 4 5 6 , 7 8 9 𝑢 C = 3 6 −2 𝑢=3 1 2 3 +6 4 5 6 −2 7 8 9 = 13 20 27

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)

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

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)

Changing Coordinates

Equation of ellipse ? Rotate 45◦ 2 2 3 3 Draw by http://web.geogebra.org/app/ ?

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

Equation of ellipse B = { 2 2 2 2 , − 2 2 2 2 }

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?

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 = 3 2 1 2 𝑏 2 = − 1 2 3 2 𝑦 𝑥′ 30◦ 𝑣 B = 𝑥′ 𝑦′ 𝑣= 𝑥 𝑦 𝑣= B 𝑣 B 𝑥 𝑥 𝑦 = 3 2 − 1 2 1 2 3 2 𝑥′ 𝑦′ − 3 𝑥 2 +2𝑥𝑦+ 3 𝑦 2 =12

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 = 3 2 1 2 𝑏 2 = − 1 2 3 2 𝑥= 3 2 𝑥 ′ − 1 2 𝑦 ′ 𝑦= 1 2 𝑥 ′ + 3 2 𝑦 ′ 𝑣 B = 𝑥′ 𝑦′ 𝑣= 𝑥 𝑦 𝑣= B 𝑣 B 𝑥 ′ 𝑦 ′ =3 𝑥 𝑦 = 3 2 − 1 2 1 2 3 2 𝑥′ 𝑦′

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

Appendix

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

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