1) Give a set-theoretic description of the given points as a subset W of R 3. a) The points on the plane x + y 2z = 0.

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1 ) Give a set-theoretic description of the given points as a subset W of R. a) The points on the plane x + y z =. x Solution: W = {x: x = [ x ], x + x x = }. x b) The points in the yz-plane. Solution: W = {x: x = [ x ], x, x R}. x ) In each case, determine whether W is a subspace of R. a) W = {x: x = x } i. θ W since = (). ii. Let u, v W, where u = [ u u ], v = [ v v ]. Then u + v = [ u + v u + v ] where u = u and v = v. Thus, u + v = u + v = (u + v ). So, the first entry in u + v is twice the second, and therefore u + v W. iii. Finally, au = a [ u u ] = [ au au ], where au = a(u ) = (au ). The first entry in au is twice the second, so therefore au W. Since θ W, u + v W, and au W, we can say that W is a subspace of R. b) W = {x: x + x = }. W is not a subspace since θ W ( + ). ) In each case, determine whether W is a subspace of R. a) W = {x: x = x x } i. θ W since = (). u v u + v ii. Let u, v W, where u = [ u ], v = [ v ]. Then u + v = [ u + v ] where u = u u u v u + v and v = v v. Thus, u + v = u u + v v = (u + v ) (u + v ). So, the third entry in u + v is twice the first minus the second. Therefore, u + v W. u au iii. Finally, au = a [ u ] = [ au ], where au = a(u u ) = (au ) au. The third u au entry in au is twice the first minus the second, so therefore au W. Since θ W, u + v W, and au W, we can say that W is a subspace of R.

2 b) W = {x: x x = x }. Solution: W is not a subspace. Consider u = [ ] and v = [ ]. Clearly, both u and v are elements 4 of W, however u + v = [ ] W since ()() 5. 5 c) W = {x: x = x, x = x } i. θ W since = () and =. u v u + v ii. Let u, v W, where u = [ u ], v = [ v ]. Then u + v = [ u + v ] where u = u, u v u + v u = u, and v = v, v = v. Thus, u + v = u + v = (u + v ), and u + v = u v = (u + v ). So, the first entry in u + v is twice the third, and the second entry is the negative of the third. Therefore, u + v W. u au iii. Finally, au = a [ u ] = [ au ], where au = a(u ) = (au ) and au = a( u ) = u au au. The first entry in au is twice the third, and the second entry is the negative of the third. Therefore au W. Since θ W, u + v W, and au W, we can say that W is a subspace of R. 4) Let U and V be subspaces of R n. Prove that the intersection, U V, is also a subspace of R n. Solution: U V = {x: x U and x V}. Let x, y U V. Then x, y U and x, y V. Now, since U and V are subspaces, x + y U and x + y V. Thus, x + y U V. Further, ax U and ax V (again since U and V are subspaces), so ax U V. Thus, U V is a subspace. 5) Given w = [ ], x = [ ], y = [ ]. Let S = {w, x, y}. Give an algebraic specification for Sp(S). Solution: The span of S is the set of all linear combinations of the vectors in S. To find an algebraic specification for Sp(S), we must solve c w + c x + c y = b. This is the same as c b solving Ac = b, where A = [w, x, y] and c = [ c ]. This gives [ c b ], which reduces b b to [ b ]. This matrix is consistent for all b R, thus Sp(S) = R. (b + b )

3 6) Give an algebraic specification for the null space and the range of the given matrix A. Solution ( st matrix): A = [ ]. For the null space, solve Ax = θ. This gives [ 6 4 ], x which means x = x and x =. Thus, N(A) = {x: x = [ x ] = x [ ], x R}. For the range, solve Ax = b. That is, reduce the matrix [ b 6 4 b ] to get [ 4b b b + b ]. Since this matrix is consistent for all b R, we can conclude that R(A) = R. For the range, you can also notice that A is a matrix, so n =. Also, notice that Null(A) = since there is only basis vector (for example, [ ] T ) in N(A). Further, since Null(A) + Rank(A) =, this means that Rank(A) =. Any vector b R(A) must be an element of R, and since Rank(A) =, we can conclude that R(A) = R. nd matrix: A = [ 7]. For the null space, solve Ax = θ. This gives [ ], which means 5 x = x =. Only θ = [ ] N(A), so N(A) = {θ}. For the range, solve Ax = b. That is, b b reduce the matrix [ 7 b ] to get [ b b ]. This matrix is only consistent if 5 b b b + b b b + b =. Thus, R(A) = {x: x = [ ], x x + x = }. x x x rd matrix: A = [ 5 4]. For the null space, solve Ax = θ. This gives [ ], which 4 means x = x = x =. Only θ = [ ] N(A), so N(A) = {θ}. Here, we have a matrix, so n =. Since Null(A) =, this means Rank(A) =. Any vector b R(A) must be an element of R, so we can conclude that R(A) = R. 7) This question is the same as Problem from section.4. You can find the detailed solution here.

4 8) For the given set S: a) Find a subset W of S that is a basis for Sp(S) using the first technique to find a basis. b) Find a basis for Sp(S) using the second technique to find a basis. Solution: For part a), the first technique to find a basis requires that we reduce the matrix [A θ], where A is the matrix with the four given vectors as its columns. This gives [ ] rref [ ] Since the matrix reduces to [I θ], there are no unconstrained variables and therefore the given set S is linearly independent and is therefore already a basis for Sp(S). For part b), reducing A T will give the identity matrix I (which is its own transpose), so another basis for Sp(S) is W = {e, e, e, e 4 }. 9) Given w = [ ], x = [ ], y = [ ]. Let S = {w, x, y}, determine whether S is a basis for R. Solution: For a matrix A = [w, x, y] and check to see if it is row equivalent to I. If it is, then the vectors in S are linearly independent, and thus for a basis for R (because there are three of them). Reducing the matrix A does indeed give I, so S is a basis for R. ) Find a basis for N(A), R(A), and give the nullity and rank of A. Solution: For the each matrix, we can find a basis for the null space by reducing the matrix [A θ]. For the first matrix, this gives [ ], so x = x x 4 and x = x + x 4. The x x 4 x + x 4 null space is N(A) = {x: x = [ x ] = x [ ] + x 4 [ x 4 ], x, x 4 R}. If we let u = [ ] and u = [ ], then a basis for N(A) is {u, u }. For a basis for the range, remember that the range is the same as the span of the vectors that make up the columns of A, so we can find a basis for the range in the same way that we find a basis for the span of a set of vectors. That is, we can either reduce the matrix [A θ] and delete the columns that correspond to unconstrained variables, or we can reduce A T to get a matrix B T and then use the nonzero columns of B for our basis. Since we have already reduced the matrix [A θ], we can see that the variables x and x 4 are unconstrained, so we can delete columns and 4 from the matrix A and

5 get a basis for the range of {[ ], [ 5]}. Since there are two vectors in the basis for the null space and in the basis for the range, this means Rank(A) = Nullity(A) =. For the second matrix, reducing [A θ] gives [I θ]. This means that the equation Ax = θ has only the trivial solution, and therefore N(A) = {θ}, so Nullity(A) =. This means Rank(A) = 4 since A is a 4 4 matrix. Since there are four columns in A, this means that the four vectors that make up the columns of A must span R 4, and therefore form a basis for R(A). ) Let W be a subspace, and let S be a spanning set for W. Find a basis for W, and calculate dim W. Solution: To find a basis, reduce the matrix [A θ], where A is the matrix whose columns are formed by the vectors in S. This gives [I θ], and so the vectors in S are linearly independent and thus form a basis for W. ) Use the Gram-Schmidt process to generate and orthogonal set from the given linearly independent vectors. Solution: For the first set of vectors, u = [ ], u = w u T w u T u u = [ ] [ ] = [ ] u = w u T w u T w u u = [ ] [ ] 6 [ ] = [ ] 6 4 For the second set of vectors, u = [ ], u = w u T w u = [ ] [ ] = [ ], and eliminating fractions gives u = [ ] u = w u T w u T w u u = [ ] [ ] 4 6 [ ] = [ ], or without fractions, u = [ ] ) This is the same as problem 5 on section.6. The solution is here. We covered the problems from section.7 in class today (7/), so see your notes for the solutions to problems,, and 4 from that section.