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4 Pages

### math171a_hw3_sol

Course: MATH 171A, Spring 2012
School: UCSD
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Word Count: 1037

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171A Math Homework 3 Solutions Instructor: Jiawang Nie February 1, 2012 1. (4 points) For a given nonzero matrix A and nonzero vector b, assume that b may be written as b = bR + bN , where bR Range(A) and bN Null(AT ). (a) Show that bR and bN are unique. Solution: By contradiction, assume there are two decompositions, i.e. b = bR + bN = ^R + ^N b b bR , ^R Range(A) and bN , ^N Null(AT ) b b and we want to show...

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171A Math Homework 3 Solutions Instructor: Jiawang Nie February 1, 2012 1. (4 points) For a given nonzero matrix A and nonzero vector b, assume that b may be written as b = bR + bN , where bR Range(A) and bN Null(AT ). (a) Show that bR and bN are unique. Solution: By contradiction, assume there are two decompositions, i.e. b = bR + bN = ^R + ^N b b bR , ^R Range(A) and bN , ^N Null(AT ) b b and we want to show that bR = ^R and bN = ^N . b b First we note that by above we have bR - ^R = ^N - bN b b since Range(A) and Null(AT ) are vector subspaces, we have that bR - ^R b ^N Null(AT ) and by the above equality we have Range(A) and bN - b bR - ^R , bN - ^N Range(A) Null(AT ) b b Now if we can just show that the only vector which lies in the intersection of Range(A) and Null(AT ) is the zero vector, we will have shown that bR = ^R and b ^N . To do so, lets suppose we have some z Range(A) Null(AT ). Since bN = b z Range(A) there exists some x such that Ax = z. Then we have, ||z||2 = ||Ax||2 = xT AT Ax = xT AT z = 0 since z Null(AT ) implies AT z = 0. Since the norm of a vector is zero if and only if the vector itself is zero, we have that z = 0. This proves our claim. (b) Show that bT bN = 0. R Solution: To show that bT bN = 0 we note that bR = Ax for some x since bR R Range(A) and thus substituting this expression for bR into the inner-product we have (Ax)T bN = xT AT bN = 0 since bN Null(AT ). 1 for which (c) If bR and bN are both nonzero, show that they are linearly independent. Solution: Suppose bR and bN are both non-zero and are linearly dependent. Then one must be a non-zero scalar multiple of the other i.e. there exists R, = 0 such that bR = bN . But by (1b) we have that 0 = bT bN = (bN )T bN = ||bN ||2 R and thus either = 0 or bN = 0. In either case we have a contradiction, so bR and bN cannot be linearly dependent i.e they are linearly independent. 2. (4 points) Consider the matrix A 7 0 0 2 7 7 6 2 A= 5 3 9 6 5 2 and vector y given by -7 -7 4 6 -7 7 14 -3 -2 0 , y = 1 . 1 4 -1 3 9 -1 1 (a) Find the rank of A. Use the Matlab command null to find a basis for Null(A). Solution: Using matlab, one should find that the rank of A is 2 and a basis for Null(A) is 0.1611 0.3879 -0.4253 -0.8017 0.7387 -0.3185 0.4795 -0.3507 Code as follows: A = [ 7 0 -7 -7; 0 2 4 6; 7 7 7 14; 6 2 -2 0; 5 3 1 4; 9 6 3 9; 5 2 -1 1]; Null(A) (b) Find the unique vectors yR Range(AT ) and yN Null(A) such that y = yR + yN . Solution: Denote the basis of Null(A) in question (a) as w1 and w2 , and use MATLAB command orth to get a basis for R(AT ), denote as z1 and z2 . Then we form the matrix B = w1 w2 z1 z2 It's full rank, and we want to get y = yR + yN , let yR = u1 w1 + u2 w2 and yN = u3 z1 + u4z2 . To get the u1 coefficients , u2 , u3, u4 , we just solve Bu = y. Then we will have y = u1 w1 + u2 w2 + u3 z1 + u4 z2 . The solution for this question is: -6.941 -0.059 -2.765 -0.235 yR = 1.412 yN = -0.412 -1.353 0.353 2 Matlab code is as follow: A = [ 7 0 -7 -7;0 2 4 6; 7 7 7 14; 6 2 -2 0; 5 3 1 4; 9 6 3 9; 5 2 -1 1]; y = [-7;-3;1;-1]; z = Null(A); w = orth(A'); B = [w z]; u = inv(B)*y; yR = w*u(1:2) yN = z*u(3:4) 3. (4 points) Suppose that the constant vector c is such that cT p 0 for all p satisfying Ap = 0. Show that this implies that cT p = 0 for all p satisfying Ap = 0. Solution: We suppose that for every vector p such that Ap = 0 we have that cT p 0. Thus, let p be any vector for which Ap = 0. Then since multiplication by A is linear operation we also have that A(-p) = -Ap = 0 which by our assumption means that cT p 0 and cT (-p) 0. This is the same as -cT p 0 or cT p 0. Since 0 cT p 0 we must have cT p = 0. 4. (4 points) Consider the equality-constrained linear program: minimize cT x subject to Ax = b, where A, b, c are given in the below A= -4 1 -3 -8 3 2 2 6 , b= -1 , 3 cT = 2 17 -1 -10 . (a) Find a basic x which is feasible for this linear program problem. Solution: It's easy to see that basic solution of constraints Ax = b will be a feasible point. So we just need to get a basic solution. Pick the {1, 2} columns of matrix A, -4 1 C= -3 -8 The determinant is nonzero, so it's invertible, solve Cy = b, we get y = (0.1429, -0.4286), and the feasible point x = (0.1429, -0.4286, 0, 0). (b) Find a minimizer x and a Lagrange multiplier , and explain whether or not they are unique. Solution: We know that since A has full row rank, all feasible vectors for which Ax = b minimize the function cT x. So take whatever you got in (4a). And this will be a minimizer, but it is certainly not unique. To find a Lagrange multiplier, solve AT = c which turns out to have a unique solution since we can show that c R(A ). This is an overdetermined linear system, it has solution, should be unique. 3 5. (4 points) Consider the equality-constrained linear program: minimize cT x subject to Ax = b. Show that if A has full column rank and b Range(A), then this linear program problem always has a unique bounded solution for any given vector c. Do Lagrange multipliers exist? If they exist, are they unique? Solution: We consider the ELP min cT x s.t. Ax = b where A has full column rank and b Range(A). These two conditions imply that the feasible region consists of one point x which is clearly bounded and x is the unique point at which cT x reaches a minimum. Moreover, AT has full row rank meaning that AT = c always possesses an infinite number of solutions. This translates to the fact that Lagrange multipliers do exist, however, they are may not unique. 4
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