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OR-1 20111 Backgrounds-Convexity  Def: line segment joining two points is the collection of points.

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Presentation on theme: "OR-1 20111 Backgrounds-Convexity  Def: line segment joining two points is the collection of points."— Presentation transcript:

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2 OR-1 20111 Backgrounds-Convexity  Def: line segment joining two points is the collection of points

3 OR-1 20112  Def: is called convex set iff whenever Convex sets Nonconvex set

4 OR-1 20113  Def: The convex hull of a set S is the set of all points that are convex combinations of points in S, i.e. conv(S)={x: x =  i = 1 k i x i, k  1, x 1,…, x k  S, 1,..., k  0,  i = 1 k i = 1}  Picture: 1 x + 2 y + 3 z, i  0,  i = 1 3 i = 1 1 x + 2 y + 3 z = ( 1 + 2 ){ 1 /( 1 + 2 )x + 2 /( 1 + 2 )y} + 3 z (assuming 1 + 2  0) x y z

5 OR-1 20114 Proposition: Let be a convex set and for, define Then is a convex set. Pf) If k = 0, kC is convex. Suppose For any x, y  kC,  Hence the property of convexity of a set is preserved under scalar multiplication. Consider other operations that preserve convexity.

6 OR-1 20115 Convex function  Def: Function is called a convex function if for all x 1 and x 2, f satisfies Also called strictly convex function if

7  Meaning: The line segment joining (x 1, f(x 1 )) and (x 2, f(x 2 )) is above or on the locus of points of function values. OR-1 20116

8 7  Def: f: R n  R. Define epigraph of f as epi(f) = { (x,  )  R n+1 :   f(x) }  Equivalent definition: f: R n  R is a convex function if and only if epi(f) is a convex set.  Def: f is a concave function if –f is a convex function.  Def: x  C is an extreme point of a convex set C if x cannot be expressed as y + (1- )z, 0 < < 1 for distinct y, z  C ( x  y, z ) (equivalently, x does not lie on any line segment that joins two other points in the set C.) : extreme points

9 OR-1 20118 Review-Linear Algebra  inner product of two column vectors x, y  R n : x’y =  i = 1 n x i y i If x’y = 0, x, y  0, then x, y are said to be orthogonal. In 3-D, the angle between the two vectors is 90 degrees. ( Vectors are column vectors unless specified otherwise. But, our text does not differentiate it.)

10 OR-1 20119  Submatrices multiplication

11 OR-1 201110  submatrices multiplications which will be frequently used.

12 OR-1 201111  Def: is said to be linearly dependent if , not all equal to 0, such that ( i.e., there exists a vector in which can be expressed as a linear combination of the other vectors. )  Def: linearly independent if not linearly dependent. In other words, (i.e., none of the vectors in can be expressed as a linear combination of the remaining vectors.)  Def: Rank of a set of vectors : maximum number of linearly independent vectors in the set.  Def: Basis for a set of vectors : collection of linearly independent vectors from the set such that every vector in the set can be expressed as a linear combination of them. (maximal linearly independent subset, minimal generator of the set)

13 OR-1 201112  Thm) r linearly independent vectors form a basis if and only if the set has rank r.  Def: row rank of a matrix : rank of its set of row vectors column rank of a matrix : rank of its set of column vectors  Thm) for a matrix A, row rank = column rank  Def : nonsingular matrix : rank = number of rows = number of columns. Otherwise, called singular  Thm) If A is nonsingular, then unique inverse exists.

14 OR-1 201113 Simutaneous Linear Equations Thm: Ax = b has at least one solution iff rank(A) = rank( [A, b] ) Pf)  ) rank( [A, b] )  rank(A). Suppose rank( [A, b] ) > rank(A). Then b is lin. ind. of the column vectors of A, i,e., b can’t be expressed as a linear combination of columns of A. Hence Ax = b does not have a solution.  ) There exists a basis in columns of A which generates b. So Ax = b has a solution.  Suppose A: m  n, rank(A) = rank [A, b] = r. Then Ax = b has a unique solution if r = n. Pf) Let y, z be any two solutions of Ax = b. Then Ay = Az = b, or Ay – Az = A(y-z) = 0. A(y-z) =  j=1 n A j (y j – z j ) = 0. Since column vectors of A are linearly independent, we have y j – z j = 0 for all j. Hence y = z. (Note that m may be greater than n.)

15  OR-1 201114

16 OR-1 201115  Operations that do not change the solution set of the linear equations (Elementary row operations)  Change the position of the equations  Multiply a nonzero scalar k to both sides of an equation  Multiply a scalar k to an equation and add it to another equation Hence X = Y. Solution sets are same.  The operations can be performed only on the coefficient matrix [A, b], for Ax = b.

17 OR-1 201116 Solving systems of linear equations (Gauss-Jordan Elimination, 변수의 치환 ) (will be used in the simplex method to solve LP problems)

18 OR-1 201117  Infinitely many solutions

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21 OR-1 201120  Elementary row operations are equivalent to premultiplying a nonsingular square matrix to both sides of the equations Ax = b

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24 OR-1 201123  So if we multiply all elementary row operation matrices, we get the matrix having the information about the elementary row operations we performed

25 OR-1 201124  Finding inverse of a nonsingular matrix A. Perform elementary row operations (premultiply elementary row operation matrices) to make [A : I ] to [ I : B ] Let the product of the elementary row operations matrices be C. Then C [ A : I ] = [ CA : C ] = [ I : B] Hence CA = I  C = A -1 and B = A -1.

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