EE 518 Homework 1 Due on Wednesday September 7 2016 Problem 1 Given a matrix A amd a vector 1 A 1 0 1 1 1 1 1 and 1 1 2 1 i Find the determinant trace transpose Row Echelon Form and rank of A ii Solve Ax iii Find the nullspace of A iv Find a lower triangular L and an upper triangular U so that A LU v Find all the eigenvalues of A and determine whether A is diagonalizable Page 1 of 5 EE 518 Homework 1 1 i det A 1 0 Due on Wednesday September 7 2016 1 1 1 1 0 2 1 trace A 1 1 1 1 1 1 0 A A 1 1 2 1 1 1 1 1 1 1 1 or A 0 A 0 1 2 0 0 0 0 12 0 1 0 1 2 Row Echelon Form 0 rank A 2 1 1 1 ii A A a 1 1 1 0 2 1 1 0 21 1 A A a 0 1 2 0 0 0 1 1 1 0 1 0 1 2 1 1 1 1 1 2 1 1 0 0 0 0 0 1 1 1 1 2 0 0 1 1 2 0 2 0 1 1 2 2 x3 is a free variable Thus x1 12 x3 12 x2 21 x3 12 x3 x3 So x 12 t 12 t 0 1 1 iii Ax 0 Thus x 1 t t 2 1 iv One possible solution is L 1 0 v I A 1 1 0 0 0 1 1 1 2 1 0 0 0 1 0 U 0 0 0 1 1 1 1 1 1 2 2 1 1 1 1 2 1 1 1 2 1 1 1 1 2 Thus the eigenvalues are 1 0 2 0 3 1 For 1 2 0 rank I A rank A 2 and rank nullspace A 1 We cannot find two linearly independent solutions for Ax 0 Thus A is not diagonalizable Problem 2 1 Determine a spanning set for nullspace A with A 1 1 1 11 1 3 x 1 t 0 v t v 0 2 3 4 3 4 5 5 2 1 1 Page 2 of 5 EE 518 Homework 1 Due on Wednesday September 7 2016 Problem 3 Let V 3 x y z for all real x y z For each of the following subsets of V determine whether or not the subset is a subspace of V If it is not a subspace show how one of the vector space requirement fails i S1 all x y z with x 2y 3z ii S2 all x y z with x 2y z iii S3 all x y z with x y 3z 5 iv S4 all x y z with x2 2y 2 z 2 i Yes ii Yes iii No because it does not contain the zero element 0 0 0 S3 iv No because it is not closed under vector addition 1 0 1 S4 and 0 1 2 S4 but 1 0 1 0 1 2 S4 Problem 4 i Find the largest possible number of independent vectors among the following vectors 1 1 1 0 0 1 0 0 1 0 v1 0 v2 1 v3 0 v4 1 v5 1 0 0 1 0 1 ii Find the largest possible number of independent matrices among the following matrices A1 1 0 0 3 A2 4 1 0 2 A3 0 1 1 4 A4 2 2 2 4 i The largest number of linearly independent vectors is 3 v1 v2 v3 are linearly independent but v4 v1 v2 and v5 v2 v3 ii The largest number of linearly independent matrices is 3 A1 A2 A3 are linearly independent but A4 2A3 Problem 5 Let vector space Pk 0 1 x k 1 xk 1 i Define two transformations T1 P3 P4 and T2 P3 P2 as follows For every polynomial p x in P3 T1 p x xp x and T2 p x p0 x dp x dx Let p1 x 1 p2 x x x2 p3 x x x2 i Show that p1 x p2 x p3 x form a basis for P3 and determine the components of p x 1 4x 2x2 relative to this basis ii Is T1 p1 x T1 p2 x T1 p3 x a spanning set for P4 Is T2 p1 x T2 p2 x T2 p3 x a spanning set for P2 Page 3 of 5 EE 518 Homework 1 Due on Wednesday September 7 2016 iii Show that T1 P3 P4 and T2 P3 P2 are linear transformations determine Ker T1 Ker T2 and their dimensions i We have that dim P3 3 and p1 x p2 x p3 x So by taking the Wronskian we have that 1 x x2 x x2 W p1 p2 p3 x 0 1 2x 1 2x 4 6 0 0 2 2 Hence p1 p2 p3 are Linearly Independent on any interval and thus they are a basis for P3 In order to find the desired components we have to solve the following equation by matching coefficients of the same power of x c1 p1 c2 p2 c3 p3 1 4x 2x2 c1 1 c2 1 c3 3 ii For the first case we have the following set x x2 x3 x2 x3 If we consider the polynomial p x 0 where 0 6 0 0 we have that p x P4 but p x span x x2 x3 x2 x3 Therefore the set is not a spanning set for P4 For the second case we have the following set 0 1 2x 1 2x If we consider an arbitrary polynomial p x 0 1 x P2 we can solve the following equation by matching coefficients of the same power of x c1 p1 c2 p2 c3 p3 0 0 x c1 0 c2 20 41 c3 20 41 Thus we can write any arbitrary polynomial in P2 as p x c1 T2 p1 x c2 T2 p2 x c3 T2 p3 x Therefore the set is a spanning set for P2 iii For T1 we have that for f g P3 and c then T1 f g x f g xf xg T1 f T1 g T1 cf xcf cxf cT1 f Thus T1 is a linear transformation So we have that Ker T1 p x P3 T1 p x 0 p x P3 xp x 0 p x P3 p x 0 0 dim Ker T1 0 For T2 we have that for f g P3 and c then T2 f g f g 0 f 0 g 0 T2 f T2 g T2 cf cf 0 cf 0 cT2 f Thus T2 is a linear transformation So we have that Ker T2 p x P3 T2 p x 0 p x P3 p0 x 0 p x P3 p x dim Ker T2 1 Problem 6 This exercise is for you to get familiar with Matlab operations For each case write the command you …