Math 2, Winter 29, Homework 7 () Find the standard matrix of the linear transformation T : R 3 R 3 obtained by reflection through the plane x + z = followed by a rotation about the positive x-axes by 6 Let S denotes the reflection and R denotes the rotation The standard matrix of reflection S through the plane x + z = can be obtained by using the formula S(u) = u 2proj n u, where n = (,, ) is the normal vector of the plane and u R 3 Since, S(e ) = (,, ) 2( (,, ) (,, ) )(,, ) = (,, ) 2 (,, ) (,, ) S(e 2 ) = (,, ) 2( )(,, ) = (,, ) 2 (,, ) (,, ) S(e 3 ) = (,, ) 2( )(,, ) = (,, ), 2 the standard matrix of S is: [S] = [S(e ) S(e 2 ) S(e 3 )] = The rotation R by π/3 about the x-axes has the standard matrix [R] = cos π/3 sin π/3 sin π/3 cos π/3 = /2 3/2 3/2 /2 Since T = R S, the required matrix is [T ] = [R S] = [R][S] = /2 3/2 3/2 /2 = 3/2 /2 /2 3/2
(2) Determine if the given linear transformation is one-to-one and/or onto Justify your answer (a) T : R 4 R 3, T (x, y, z, w) = (x 4y + 8z + w, 2y z + 3w, 5w) (b) T : R 3 R 4, T (x, y, z) = (2x + 3y, 3x + y + z, x z, y z) (a) We first observe that the standard matrix [T ] of T is [T ] = 4 8 2 3 5 The matrix [T ] is already in an echelon form, and we see that each row of [T ] has a pivot position, therefore T is onto Also, not every column of T has a pivot position, that is, T is not one-to-one (b) We have [T ] = 2 3 3 4 Thus, each column of [T ] has a pivot position, hence T is one-to-one On the other hand, not every row of [T ] has a pivot position, that is, T is not onto (3) u = (, 2), v = (3, 5) and w = (7, 5) be vectors in R 2 (a) Express w as a linear combination of u and v, that is, find scalars α and β such that w = α u + β v (b) Let T : R 2 R 2 be the linear transformation such that T ( u) = ( 4, 3) and T ( v) = (7, 8) Find T ( w) (a) The vector equation α u + β v = w, that is, α(, 2) + β(3, 5) = (7, 5), is equivalent to the linear system { α + 3β = 7 2α + 5β = 5 Solving the system we have α = 2, β = 9 Thus, w = ( 2) u + 9 v (b) Since T is a linear transformation, T ( w) = T (α u+β v) = αt ( u)+βt ( v) = ( 2)( 4, 3)+9(7, 8) = (43, 2)
(4) Determine whether w lies in Span{u, v} (a) u = (, 2, 2), v = (, 3, ), w = (3,, 5) (b) u = (2,, 3, 5), v = (, 2, 2, 3), w = (2, 5, 8, 3) (a) w Span{u, v} x u + x 2 v = w, x, x 2 R The last vector equation is equivalent to the linear system whose augmented 3 matrix is: [ u v w] = 2 3 An echelon form of 2 5 2 5 this matrix is 2 6 Therefore, the system is consistent ( it has the unique solution, x = 4, x 2 = 3) and w is in Span{u, v} (b) As in the part (a): w Span{u, v} x u + x 2 v = w, x, x 2 R The last vector equation is equivalent to the linear system whose augmented matrix is: [ u v w] = 2 5 2 2 3 2 8 An echelon form of 5 3 3 this matrix is and w / Span{u, v} 2 5 5 8 8 23 9 Therefore, the system is inconsistent (5) Under what condition(s) does b = (b, b 2, b 3 ) lie in the span of S = {(, 2, ), (6, 4, 2)}? Does w = (9, 2, 7) lie in Span{S}? What about w = (4,, 8)? Explain your answer Let u = (, 2, ), v = (6, 4, 2) Then, as in the previous question, b Span{u, v} x u + x 2 v = b, x, x 2 R
The last vector equation is equivalent to the linear system whose augmented 6 b matrix is: [ u v b] = 2 4 b 2 Using row operations, 2 b 3 6 b 6 b 6 b [ u v b] = 2 4 b 2 = 2 8 b 2 2b = 2 8 b 2 2b 2 b 3 8 b 3 + b b + b 2 + b 3 Thus, b = (b, b 2, b 3 ) lies in the span of S if and only if b + b 2 + b 3 =, ( ) which is an equation of a plane in R 3 through the origin The vector w = (9, 2, 7) satisfies the condition ( ), hence lies in Span{S}, while the vector w = (4,, 8) does not satisfies the condition ( ), hence, does not lie in Span{S} (6) Determine if the following vectors span R 3 or not: {(2,, ), (,, ), (, 2, ), (2, 2, )} The set {(2,, ), (,, ), (, 2, ), (2, 2, )} spans R 3 means that for any vector b = (b, b 2, b 3 ) in R 3, the vector equation x (2,, ) + x 2 (,, ) + x 3 (, 2, ) + x 4 (2, 2, ) = (b, b 2, b 3 ) has a solutionthis vector equation is equivalent to the linear system whose augmented matrix is 2 2 b 2 2 b 2 An echelon form of this matrix is b 3 b 3 3 2 b 2 + b 3 Since no values of b, b 2, b 3 can produce 8 8 b + 3b 2 + b 3 a row of the form ( ), (where a non-zero number), the system is consistent Hence, the vectors span R 3 (7) Let v = (, 3,, ), v 2 = (6,, 5, ), v 3 = (4, 7,, c), where c is a constant (a) Find value(s) of c so that {v, v 2, v 3 } is linearly dependent in R 4 (b) For each value of c found in (a), find a nontrivial linear dependence relation among v, v 2 and v 3
(a) The set {v, v 2, v 3 } is linearly dependent if and only if the vector equation x v +x 2 v 2 +x 3 v 3 = has a nontrivial solution, that is, if and only if the linear system whose augmented matrix is [v v 2 v 3 ] has a nontrivial solution An echelon form of [v v 2 v 3 ] = is 5 3 2 c 3 {v, v 2, v 3 } is linearly dependent 6 4 3 7 5 c Hence, for c = 3, there is a free variable and (b) For c = 3, the reduced row echelon form of [v v 2 v 3 ] is 7/3 2/3 Therefore, x = 7 3 x 3, x 2 = 2 3, where x 3 is free For example, if x 3 = 3, then x = 7 and x 2 = 2 Thus, a nontrivial linear dependence relation among v, v 2, v 3, is 7v 2v 2 + 3v 3 = (8) Suppose {v, v 2 } are linearly independent vectors in R 4 Given w = v 3v 2, w 2 = v + v 2 and w 3 = 2v + v 2, show that {w, w 2, w 3 } is linearly dependent set The set {w, w 2, w 3 } is linearly dependent set if and only if the vector equation x w + x 2 w 2 + x 3 w 3 = has a nontrivial solution, that is, if and only if the linear system whose augmented matrix is [w w 2 w 3 ] has a nontrivial solution We have x w +x 2 w 2 +x 3 w 3 = x (v 3v 2 )+x 2 (v +v 2 )+x 3 ( 2v +v 2 ) = (x + x 2 2x 3 )v + ( 3x + x 2 + x 3 )v 2 = Since the set {v, v 2 } is linearly independent, only trivial linear combination of v, v 2 is equal, that is { x + x 2 2x 3 = 3x + x 2 + x 3 = This homogeneous system must have a nontrivial solution ( more unknown than equations) Therefore, the set {w, w 2, w 3 } is linearly dependent set Alternative We can show that {w, w 2, w 3 } is linearly dependent set, by expressing one of w, w 2, w 3 as a linear combination of the
other two Since w 2 = v + v 2 and w 3 = 2v + v 2, by subtracting them, we get, w 2 w 3 = 3v, that is, v = 3 w 2 3 w 3 Plugging this into the first equation, we get, w 2 = 3 w 2 3 w 3 + v 2 Solving for v 2 gives v 2 = 2 3 w 2 + 3 w 3 Finally, w = v 3v 2 = 3 w 2 3 w 3 3( 2 3 w 2 + 3 w 3) = ( 5 3 )w 2 4 3 w 3 (9) Let A = 4 5 6 9 3 2 4 2 2 3 5 7 8 a basis for Nul(A), the nul space of A, 2 a basis for Col(A), the column space of A, 3 a basis for Row(A), the row space of A, 4 rank(a), 5 nullity(a) The reduced echelon form of A is B = Find the following: 2 2 The augmented matrix [A ] of the linear system Ax =, where x = [x, x 2, x 3, x 4, x 5 ] T and = [,,, ] T has the reduced row echelon form [B ] Thus, x and x 2 are leading variables and x 3, x 4, x 5 are free variables Therefore, x is in Nul(A) means that x = x x 2 x 3 x 4 x 5 Hence, = x 3 2x 4 x 5 x 3 x 4 2x 5 x 3 x 4 x 5, 2, 2 = x 3 +x 4 2 is a basis of Nul(A) +x 5 2
2 The pivot columns of B, that is, columns of B which contain a leading one, are columns and 2 Thus, columns and 2 of A form a basis of 4 Col(A) Therefore, 3, 2 ia a basis of Col(A) 2 3 3 Nonzero rows of B, that is {(,,, 2, ), (,,,, 2)}, form a basis of Row(B) = Row(A) 4 Rank(A) = dim(col(a)) = 2 5 Nullity(A) = dim(nul(a)) = 3 () Let v = (, 2, ), v 2 = (2, 9, ), v 3 = (3, 3, 4) (a) Show that B = {v, v 2, v 3 } form a basis of R 3 (b) Find the coordinate vector (w) B of w relative to B if w = (5,, 9) 2 3 (a) Let A = [v v 2 v 3 ] = 2 9 3 Then det(a) = Thus, 4 vectors v, v 2, v 3 are linearly independent Since dim(r 3 ) = 3, by the dimension argument (every three linearly independent vectors in a threedimensional vector space form a basis, see Theorem 545, Section 54 of the text ), B = {v, v 2, v 3 } form a basis for R 3 (b) The coordinate vector (w) B of w relative to B is a vector (x, x 2, x 3 ) such that x v + x 2 v 2 + x 3 v 3 = w This vector equation is equivalent to the linear system whose augmented matrix is [v v 2 v 3 w] = 2 3 5 2 9 3 The system must have the unique solution ( since 4 9 B is a basis of R 3 ) The solution is x =, x 2 =, x 3 = 2 Therefore,(w) B = (,, 2)
() Let S = {v, v 2, v 3, v 4, v 5 }, where v = (,, 5, 2), v 2 = ( 2, 3,, ), v 3 = (4, 5, 9, 4), v 4 = (, 4, 2, 3), v 5 = ( 7, 8, 2, 8) (a) Find a subset B of S that form a basis for the subspace W = Span{v, v 2, v 3, v 4, v 5 } (b) For each vector in S that is not in B find the coordinate vector relative to B (a) Let A = [v v 2 v 3 v 4 v 5 ] = 2 4 7 3 5 4 8 5 9 2 2 2 4 3 8 Then W = Span{v, v 2, v 3, v 4, v 5 } = Col(A) The reduced echelon 2 form of A is the matrix B = 3 2 = [v v 2 v 3 v 4 v 5] We see that the basis for Col(B) is {v, v 2, v 4} Thus, a basis B for W = Span{v, v 2, v 3, v 4, v 5 } = Col(A) is B = {v, v 2, v 4 } (b) By inspection of columns of B we see that { v 3 = 2v v 2 v 5 = v + 3v 2 + 2v 4 Since row operations preserve linear dependence relations we have { v3 = 2v v 2 v 5 = v + 3v 2 + 2v 4 Therefore, coordinate vectors of v 3 and v 5 relative to the basis B are: (v 3 ) B = (2,, ), (v 5 ) B = (, 3, 2) (2) Find a vector v in R 4 so that {(,,, ), (,,, ), (2,,, ), v} is a basis of R 4 Vectors v = (,,, ), v 2 = (,,, ), v 3 = (2,,, ) are linearly independent Thus, the set {v, v 2, v 3 } can be extended to a basis of R 4 2 2
So,we can take v = e 2 = (,,, ), hence, {v, v 2, v 3, e 2 }, is a basis for R 4 Note that this is not the unique answer (3) Suppose that A is 4 5 matrix and b is a column vector in R 4 Given the following data, determine whether each of the 6 systems of equations Ax = b has (a) no solution (b) one solution (c) infinitely many solutions (d) the data do not give enough information to tell (e) the data are impossible rank(a) rank([a b]) (i) 4 5 (ii) 4 4 (iii) 4 3 (iv) 3 4 (v) 3 5 (vi) 3 3 Briefly explain your answers Here [A b] denotes the augmented matrix of Ax = b rank(a) rank([a b]) (i) 4 5 impossible, maximal rank of [A b] is 4 (ii) 4 4 infinitely many solutions, one free variable (iii) 4 3 impossible, rank(a) rank([a b]) (iv) 3 4 no solution, rank(a) rank([a b]) (v) 3 5 impossible (vi) 3 3 infinitely many solutions, 2 free variables