Linear Algebra: Understanding Vectors, Matrix Operations, and Coordinate Systems, Slides of Advanced Computer Programming

Download Linear Algebra: Understanding Vectors, Matrix Operations, and Coordinate Systems and more Slides Advanced Computer Programming in PDF only on Docsity! Lecture 06: Linear Algebra, A Review, Part I Docsity.com  Linear Algebra that you should know…  3D Coordinate Geometry  3D Points and Vectors  dot products and cross products  Vector and Matrix notations and algebra  Properties of matrix multiplication (associative property, but NOT communicative property)  Associative => (5 + 2) + 1 = 5 + (2 + 1)  Commutative => 5 +2 + 1 = 1 + 2 + 5  Matrix operations (multiplication, transpose, inverse, etc.) Linear Algebra for 3D Graphics Docsity.com Vectors b Denotes: >» Magnitude » Direction >» NO POSITION!! > In R¢, a vector can be defined as an ordered d-tuple: Vy ee ee net V2 | v= : 2s Vd : = » Avector v is often written as v ( for clarity ® Docsity.com — co i Vectors >» Because a vector has no position, one way to think of Vy a 2D vector v = E | isasan y “offset from the origin’. > Such offset can be translated (moved around) Docsity.com  Find the length of the following vector: Vectors - Length Docsity.com Vectors — Scalar Multiple > Examples of scalar multiples of the vector v = A r > These vectors are said to be parallel to the vector v. > Question: What happens if a = 0? ® Docsity.com Vectors — Unit Vector If I don’t care about magnitude of the vector - that is, all I care about is the direction, I can represent the vector v such that: ||v|] = 1 > This vector v is said to be a “unit vector” (often denoted as 7, called v-hat), and the process of making a vector be of length = 1 is called “normalization”. ® Docsity.com Vectors — Unit Vector > Question: Given a vector v, how do you find 0? > Question: Can every vector | s be normalized? | e ® Docsity.com  Vectors – Addition 6 7 8 9 10 6 7 8 9 10 1 2 3 5 4 1 2 3 4 5 v u v + u 6 7 8 9 10 6 7 8 9 10 1 2 3 5 4 1 2 3 4 5 v u u + v Docsity.com  Vectors - Subtraction 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4 5 -4 -3 -2 -1 v u -u -u Docsity.com  Vectors – Basis 1.2 1.4 1.6 1.8 2 1.2 1.4 1.6 1.8 2 0.2 0.4 0.6 1 0.8 0.2 0.4 0.6 0.8 1 Docsity.com  Vectors – Basis 6 7 8 9 10 6 7 8 9 10 1 2 3 5 4 1 2 3 4 5 Docsity.com Vector - Basis > More formally, given a 2D vector w, we can express it using two basis vectors u and v: w= [fi] = aut bv = a fs] +0 [3] = [oa 2 ea > Question: Can you think of two vectors that CANNOT form basis vectors for a 2D space? > (Hint: think linear independence) ® Docsity.com  Vector - Basis OK OK not OK not OK Docsity.com Vector — Cross Product > The Cross Product of two vectors is a vector that is perpendicular to both original vectors. That is, it is normal to the plane containing the two vectors. Vy U4 > Given two vectors v = a and u = a , the dot V3 U3 product is written as v X u. ® Docsity.com Vector — Cross Product >» Important Note: > Cross Product is not commutative. That is: vxu #UuUXV > In fact, vxu=—(uxv) axb > The “Right Hand Rule” ‘ ® Docsity.com Vector — Cross Product > Vx > More formally: the cross product of v = bs and Vz Ux Vy Ux u = |Uy| is written as: v X u = s X |Uy |, and is uz Vz uz defined as: Vy Ux VyUz — VzUy Vy x Uy = | VZUy — Vx Uz Vz Uz ® Docsity.com  Deriving the cross product… For a simple 2D Case: Vector – Cross Product Docsity.com Vector — Cross Product >» When the axes are x and y, we get: areary,y) = A,By — B,Ay > We could have easily labeled the axes as y and z, or z and x, and we would get: areary,z) = AyB, — ByA; aredarz,x) = A,B, — BA > In other words, if we take two 3D vectors, and project them down to the (x,y), (y,z), and (z,x) planes, we could get the areas of the parallelogram in each of the planes. ® Docsity.com Vector — Cross Product > Since we can think of the (x,y) plane as defined by a 0 unit vector Z = [| at the origin (0, 0, 0). Meaning 1 that we can think of the relationship as: Z = X x 9, we can rewrite the previous equations as: area, y) = (AyBy — ByAy)Z > Similarly: aredyy7) = (AyB, — ByAz)xX areaczx) = (A,B, — BrAy)V ® Docsity.com Vector — Cross Product > Vy Ux > Given two v = |Vy| and u = |Uy|, we can rewrite Vz uz each as a sum of its components in the i, j, k basis, such that: v = (v,xi + vyj + vk), and u = (uxt + uyj + uzk) >Sovxu= (v,i + vyj + v,k) x (ui + uyj + u,k) = VyUxl XUF VyUyl XJ tee = (vyu, — vzUy )it (vu, — VyUz)j + (Vy~Uy — VyU, )k Docsity.com Vector — Cross Product >» One More Thought: > For those of you who are ninja’s in linear algebra. You might have already noticed that: ij ok vxXu=det|% Vy WY > Recall that the determinant of a matrix is: VyUzl — VzUyl + VzUyj — VyUzj + VyUyk — vyuyk Docsity.com Vector — Cross Product > Recall that the determinant of a matrix is: VyUzl — VzUyl + VzUyj — VyUzj + VyUyk — vyuyk > Organize the terms a little bit, and you get: Vy Ux VyUz — VzUy ® Docsity.com Vector — Dot Product > As it turns out, if u and v are two non-zero vectors: usv = |lullllvl| cos(@) ® Docsity.com Vector — Dot Product } This is important because if I want to know the angle between two vectors, I can trivially compute: uv = |lulll|vl] cos(@) uUu:v cos(@) = () = Toile 6 = cos 1( ui? ) ello >» Note that if both wu and v are unit vectors, then: 6 = cos 1(u:v) > Note that 6 will be in radians (that is, 0 < 0 < 7) Docsity.com  Other useful things with dot products  What are the results? Vector – Dot Product Docsity.com  Vector – Dot Product 1 2 3 4 5 1 2 3 4 5 -5 -4 -3 -1 -2 Docsity.com Vector — Dot Product > More use of the Dot Product: Finding the length of a projection > If wis a unit vector, then v - u is the length of the projection of v onto the line containing u Recall dot product: v:u=|v||u| cos @ If |u| = 1, then v:u=|v|cos@ = |u’| Docsity.com  Vector – Dot Product |v| sin(θ) |v| cos(θ) radius = ||v|| Docsity.com