Math 414 Lecture 2 Everyone have a laptop?

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Math 44 Lecture 2 Everyone have a laptop? THEOREM. Let v,...,v k be k vectors in an n-dimensional space and A = [v ;...; v k ] v,..., v k independent v,..., v k span the space v,..., v k a basis v,..., v k independent v,..., v k span the space v,..., v k is a basis ˆ k < n ˆ n < k ˆ k = n rank(a) = k rank(a) = n k = rank(a) = n `Classify the set u = [, 2, 3], v = [4, 5, 6], w = [7, 8, 9] as spanning/nonspanning, dependent/independent. 23 Solution. A = 456 rref(a) = 2 789 rank(a) =? Are u, v, w dependent? Do u, v, w span the space R 3?

`Write b = [,,] as a linear combination of u, v, w. First transpose u, v, w and b. Let A = [ u, v, w]. Thus u 2 3, v 4 5 6, w 7 8 9, b b is a linear combination of u, v, w iff xu + yv + zw = b for some x, y, z iff A*[x; y; z] = b has a solution., A [ A b ] = rref([a b] ) = x y z x y z 2 3 4 5 6 7 8 9 Basic solution: x = -ƒ, y = ƒ, z =. - 2 4 7 2 5 8 3 6 9 Thus [ ] = -ƒ[ 2 3] + ƒ[4 5 6]. Check this. -ƒ ƒ

`Write b = [,, ] as a linear combination of u, v, w. [A b] = rref( [A b]) x y z x y z 2 3 4 5 6 7 8 9 Is b = [,, ] a linear combination of u, v, w? - 2 `Given points x, y ir n, plot x y, 2 3x 3y, 2x 2y, What is the sum of the coefficients? 3x 2 3y, x y LEMMA. The line segment xy between x and y = the set of points between x and y = points of the form ax + by where a +b = and a, b >. = the convex combinations of x and y.

DEFINITION. For any subset S of R n : S is convex iff x, y i S ˆ all points between x & y are in S. S is closed iff it contains all points on its boundary. S is unbounded iff it has points arbitrarily far apart. iff it has points arbitrarily far from the origin. S is bounded otherwise. An extreme point (or vertex) of S is a point of S which is not between two other points of S. `Mark the following as convex, closed, or Bounded. Circle the extremes.

LEMMA. The intersection of convex sets is convex. The set of solutions to a linear equation is a line in R 2, a plane in R 3, a hyperplane in R n for n > 3. Each divides its space into two halves. Solutions for a non-strict (i.e., < or >) linear inequality additionally includes one of the these halves. DEFINITIONS. A closed half-space is the set of solutions to a non-strict (i.e., < or >) linear inequality. E.g. y x. Closed half-spaces are closed and convex, hence so is any intersection of such spaces. An intersection of finitely many closed half-spaces is called a convex polyhedron. In R 2, an inequality determines a half-plane. The half plane lies on the origin side of the line iff the origin satisfies the inequality iff the inequality is true when x = and y =.

`Graph the convex polyhedra. () Graph the equalities via their intercepts. (2) Draw arrows indicating the directions of the half planes (check if the origin satisfies the inequality). (3) Fill in the polyhedra. (4) Circle and label the extremes. `Graph the convex polyhedra. a: x +y < x >, y > `Graph the convex polyhedra. a: y- x < b: y + x > x >, y >

`Given points x, y, z ir n, plot x y z, 2x 4y 4z 3 x 3 y 3 z, 2 x 2 y z, DEFINITIONS. A convex combination of points x,..., x n is a point a x + a 2 x 2 +...+ a n x n where a +...+ a n = and a >,..., a n >. The convex hull of a set X of points = the set of all convex combinations of points from X = the smallest convex set containing X. The convex hull of a finite set is called a convex polytope. `xy is the convex hull of x and y.

THEOREM. (a) A closed bounded convex set is the convex hull of its extreme points. For closed bounded convex sets S, (b) S is a convex polytope iff S has only finitely many extreme points iff S is a convex polyhedron. (c) In a polytope, every point is a convex combination of extremes. The set of extremes is like a basis with linear combination replaced by convex combination. For a basis, every point is a linear combination of the basis but the basis elements are not linear combinations of other basis elements. For extremes, every point is a convex combination of extremes but extremes are not convex combinations of other points.

THEOREM. If f is a linear function on a closed bounded convex set, then every local minimum (or maximum) is also an absolute minimum (or maximum). If f is not constant, the maxima (minima) lie on the boundary. b

THEOREM. For linear functions on nonempty closed bounded convex sets: (a) Absolute maximas and minimas always exist. (b) Local maximas and minimas are absolute maximas and minimas. PROOF. (a) Linear functions are continuous and continuous functions on closed bounded sets always have absolute maxima and minima. (b) Suppose f (a) = r is a local maximum and suppose f (b) = s is a higher absolute max. Since f is linear and its domain is convex, its graph includes the line through (a, r) and (b, s). But then f (a) = r can t be a local max. f(b)=s f(a)=r X X X X a b

For linear functions, local and absolute maximas are the same, so we ll just say maxima. DEFINITION. For a linear function f (x, y) = ax + by, the gradient vector of f (x, y) is the vector [a, b]. Note: f (x, y) = a ax + by = x, y. b The gradient points in the direction of maximum increase. An extreme point has the largest projection onto the gradient line. `Suppose f (x, y) = y. On each polytope, find where f (x, y) = y has a maximum and a minimum. Gradient Diamond Square Wedge Do the same for f x, y x y.

THEOREM. For linear functions on closed bounded convex sets, the maxima are: (a) a single extreme point of maximum value, (b) the convex hull of extreme points of max value, (c) empty (happens only when the convex set is Q).