ECS130 Scientific Computing. Lecture 1: Introduction. Monday, January 7, 10:00 10:50 am
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1 ECS130 Scientific Computing Lecture 1: Introduction Monday, January 7, 10:00 10:50 am
2 About Course: ECS130 Scientific Computing Professor: Zhaojun Bai Webpage:
3 Today s Agenda Mathematics Review: Linear Algebra
4 Vector spaces over R Denote a (abstract) vector by v. A vector space which satisfies V = {a collection of vectors v} All v, w V can be added and multiplied by a R: v + w V, a v V The operations +, must satisfy the axioms: For arbitrary u, v, w V, 1. + commutativity and associativity: v + w = w + v, ( u + v) + w = u + ( v + w). 2. Distributivity: a( v + w) = a v + a w, (a + b) v = a v + b v, for all a, b R identity: there exists 0 V with 0 + v = v inverse: for any v V, there exists w V with v + w = identity: 1 v = v. 6. compatibility: for all a, b R, (ab) v = a (b v).
5 Example Euclidean space: R n = { a (a 1, a 2,..., a n ): a i R }. Addition: (a 1,..., a n ) + (b 1,..., b n ) = (a 1 + b 1,..., a n + b n ) Multiplication: c (a 1,..., a n ) = (ca 1,..., ca n ) Illustration in R 2 : b a + b 2 a a a
6 Example Polynomials: { R[x] = p(x) = i a i x i : a i R }. Addition and multiplication in the usual way, e.g. p(x) = a 0 + a 1 x + a 2 x 2, q(x) = b 1 x: Addition: p(x) + q(x) = a 0 + (a 1 + b 1 )x + a 2 x 2. Multiplication: 2p(x) = 2a 0 + 2a 1 x + 2a 2 x 2.
7 Span of vectors Start with v 1,..., v n V, and a i R, we can define v n a i v i = a 1 v 1 + a 2 v a n v n, i=1 Such a v is called a linear combination of v 1,..., v n. For a set of vectors S = { v i : i I}, all its linear combinations define { } span S a i v i : v i S and a i R i
8 Example in R 2 Observation from (c): adding a new vector does not always increase the span.
9 Linear dependence A set S of vectors is linearly dependent if it contains a vector k v = c i v i, for some v i S\{ v} and nonzero c i R. i=1 Otherwise, S is called linearly independent. Two other equivalent defs. of linear dependence: There exists { v1,..., v k } S\{ 0} such that k c i v i = 0 where c i 0 for all i. i=1 There exists v S such that span S = span(s\{ v}).
10 Dimension and basis Given a vector space V, it is natural to build a finite set of linearly independent vectors: { v 1,..., v n } V. The max number n of such vectors defines the dimension of V. Any set S of such vectors is a basis of V, and satisfies span S = V.
11 Examples The standard basis for R n is given by the n vectors Since e i = (0,..., 0, 1, 0,..., 0) for i = 1,..., n }{{}}{{} i 1 n i ei is not linear combination of the rest of vectors. For all c R n, we have c = n i=1 c i e i. Hence, the dimension of R n is n. A basis of polynomials R[x] is given by monomials {1, x, x 2,... }. The dimension of R[x] is.
12 More about R n Dot product: for a = (a 1,..., a n ), b = (b 1,..., b n ) R n Length of a vector a 2 = a b = Angle between two vectors n a i b i. i=1 a a 2 n = a a. θ = arccos a b a 2 b 2. (*Motivating trigonometric in R 3 : a b = a 2 b 2 cos θ.) Vectors a, b are orthogonal if a b = 0 = cos 90.
13 Linear function Given two vector spaces V, V, a function L: V V is linear, if it preserves linearity. Namely, for all v 1, v 2 V and c R, L[ v1 + v 2 ] = L[ v 1 ] + L[ v 2 ]. L[c v1 ] = cl[ v 1 ]. L is completely defined by its action on a basis of V: L[ v] = i c i L[ v i ], where v = i c i v i and { v 1, v 2,... } is a basis of V.
14 Examples Linear map in R n : L: R 2 R 3 defined by L[(x, y)] = (3x, 2x + y, y). Integration operator: linear map L: R[x] R[x] defined by L[p(x)] = 1 0 p(x)dx.
15 Matrix Write vectors in R m in column forms, e.g., v 1 = v 11. v m1, v 2 = v 12. v m2,..., v n = v 1n. v mn. Put n columns together we obtain an m n matrix v 11 v v 1n V v 1 v 2... v n v 21 v v 2n =.... v m1 v m2... v mn The space of all such matrices is denoted by R m n.
16 Unified notation: Scalars, Vectors, and Matrices A scalar c R is viewed as a 1 1 matrix c R 1 1. A column vector v R n is viewed as an n 1 matrix v R n 1.
17 Matrix vector multiplication A matrix V R m n can be multiplied by a vector c R n : c 1 v 1 v 2... v n. = c 1 v 1 + c 2 v c n v n. c n Elementwisely, we have v 11 v v 1n c 1 c 1 v 11 + c 2 v c n v 1n v 21 v v 2n c = c 1 v 21 + c 2 v c n v 2n.. v m1 v m2... v mn c n c 1 v m1 + c 2 v m2 + + c n v mn
18 Using matrix notation Matrix vector multiplication can be denoted by A }{{} R m n }{{} x = }{{} b R n R m. M R m n multiplied by another matrix in R n k can be defined as M[ c 1,..., c k ] [M c 1,..., M c k ].
19 Example Identity matrix I n e 1 e 2... e n = It holds I n c = c for all c R n.
20 Example Linear map L[(x, y)] = (3x, 2x + y, y) satisfies 3 0 [ ] 3x L[(x, y)] = 2 1 x = 2x + y. y 0 1 }{{} y }{{} R }{{} 2 R 3 2 R 3 All linear maps L: R n R m can be expressed as L[ x] = A x, for some matrix A R m n.
21 Matrix transpose Use A ij to denote the element of A at row i column j. The transpose of A R m n is defined as A T R n m (A T ) ij = A ji. Example: 1 2 A = 3 4 A T = 5 6 [ ] Basic identities: (A T ) T = A, (A + B) T = A T + B T, (AB) T = B T A T.
22 Examples: Matrix operations with transpose Dot product of a, b R n : n a b = a i b i = [ b ] 1 a 1... a n. = a T b. i=1 b n Residual norms of r = A x b: A x b 2 2 = (A x b) T (A x b) = ( x T A T b T )(A x b) = b T b b T A x x T A T b + x T A T A x (by b T A x = x T A T b) = b b T A x + A x 2 2.
23 Computation aspects Storage of matrices in memory: 1 2 Row-major: Column-major: Multiplication b = A x for A R m n and x R n : Access A row-by-row: 1: b = 0 2: for i = 1,..., m do 3: for j = 1,..., n do 4: b i = b i + A ij x j 5: end for 6: end for Access column-by-column: 1: b = 0 2: for j = 1,..., n do 3: for i = 1,..., m do 4: b i = b i + A ij x j 5: end for 6: end for
24 Linear systems of equations in matrix form Example: find (x, y, z) satisfying 3x + 2y + 5z = 0 4x + 9y 3z = x y = 0 7 2x 3y 3z = z 1 Given A = [ a 1,..., a n ] R m n, b R m, find x R n : A x = b. Solution exists if b is in column space of A: { n } b col A {A x: x R n } = x i a i : x i R. i=1 The dimension of col A is defined as the rank of A.
25 The square case Let A R n n be a square matrix, and suppose A x = b has solution for all b R n. We can solve The inverse satisfies (why?) A x i = e i, for i = 1,..., n. A [ ] x 1 x 2... x n = I n }{{} A 1 AA 1 = A 1 A = I n and (A 1 ) 1 = A. Hence, for any b, we can express the solution as x = A 1 A x = A 1 b.
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