That is W = f • x y ‚: xy ‚ 0g. a. If u is in W and c is any scalar, is cu in W? Why? b. Find speciﬁc vectors u and v in W such that u+v is not in W. ...

0 downloads 13 Views 80KB Size

No documents

4.1.2. Problem Restatement: · Let ¸ W be the union of the first and third quadrants in the xyx plane. That is W = { : xy ≥ 0}. y a. If u is in W and c is any scalar, is cu in W ? Why? b. Find specific vectors u and v in W such that u + v is not in W . This is enough to show that W is not a vector space. Final Answer: a. cu is in W for all u in W and scalars c. Proof in work below. · ¸ · ¸ 0 −1 b. Let u = and v = , say, but there are many other possibilities. 1 0 · ¸ · ¸ x cx Work: a. Let u = ∈ W and c ∈ R. cu = so cu ∈ W just in case cxcy ≥ 0. But y cy xy ≥ 0 since u ∈ W . Also cxcy = c2 xy. Since c2 ≥ 0, it follows that c2 xy ≥ 0. Therefore, cxcy · ≥ ¸0 as required; · ¸ that is, cu ∈ W . · ¸ · ¸ · ¸ 0 −1 0 −1 −1 b. and are each in W , but + = is not in W . 1 0 1 0 1 4.1.6. Problem Restatement: Determine if the set {p(t)|p(t) = a + t2 , a ∈ R} is a subspace of P2 . Justify. Final Answer: The set {p(t)|p(t) = a + t2 , a ∈ R} is not a subspace of P2 since the zero vector 0 = 0 + 0t + 0t2 is not in the set. Work: None required.

2t 4.1.10. Problem Restatement: Show the set H of all vectors of the form 0 is a subspace −t 3 of R by finding a spanning set. 2 Final Answer: H = Span{ 0 }, so H is a subspace by Theorem 1. −1 Work: None required.

1

−a + 1 4.1.16. Problem Restatement: Let W be the set of all vectors of the form a − 6b , where 2b + a a and b are arbitrary scalars. Either find a set of vectors S spanning W or give a counter example to show W is not a vector space. Final Answer: W is not a subspace because 0 ∈ / W. −a + 1 0 a − 6b 0 . Then a = 1, so a − 6b = 0 implies b = 16 whereas = Work: Suppose 2b + a 0 2b + a = 0 implies a contradictory result that b = − 12 . Other counter examples are possible. 4.1.32. Problem Restatement: T Let H and K be subspaces of a vector space V . The intersection T of H and K, written H K, is the set of v ∈ V such that v ∈ H and v ∈ K. Show H K is a subspace of V . Give an example in R2 to show that the union of two subspaces is not, in general, a subspace. T Final Answer: To prove H K is a subspace of V we must show it contains 0 and is closed under addition and multiplication· by ¸scalars. This is · done ¸ in the work section below where 1 0 we also show the union of Span{ } and Span{ } is not a subspace of R2 . 0 1 T Work: 1. (Zero) Since 0 ∈ H and 0 ∈ K it follows that 0 ∈ H K. T 2. (Addition) Let u, v ∈ H K. Then u, v ∈ H so u + v ∈ H since H is closed under addition. Also, T u, v ∈ K so uT+ v ∈ K since K is closed under addition. Therefore, u + v ∈ H K. Therefore, H K is closed under addition. T 3. (Scalar multiplication) Let u ∈ H KTand let c ∈ R. u ∈ H, soT cu ∈ H. As well, u ∈ K, so cu ∈ K. Therefore, cu ∈ H K and it follows that H K is closed under multiplication by scalars. · ¸ · ¸ S 1 0 4. (Union example) Let W = Span{ } Span{ }. W is not closed under addi0 1 · ¸ · ¸ · ¸ · ¸ · ¸ 1 0 1 0 1 tion since ∈ W and ∈ W , but + = ∈ / W. 0 1 0 1 1

2

Section 4.2 Null, Column Spaces and Linear Transformations

5 5 21 19 4.2.2: Problem Restatement: Determine if w = −3 is in N ul(A), where A = 13 23 2 . 2 8 14 1 Final Answer: w ∈ N ul(A) 0 25 − 63 + 38 65 − 69 + 4 = 0 . Work: Aw = 40 − 42 + 2 0

b − 5d 2b 4.2.12: Problem Restatement: Let W = { 2d + 1 d theorem to show W is a vector space, or find a Final Answer: W b − 5d 2b Work: If 2d + 1 d this value for d

: b, d ∈ R}. Either use an appropriate specific example to the contrary.

is not a vector space since it does not contain 0. 0 = 0 , then d = 0 from the forth coordinate. But upon substituting 0

into the third coordinate, we get 1 = 0, a contradiction. · ¸ 1 3 5 0 . Find a nonzero vector in each of 4.2.22: Problem Restatement: Let A = 0 1 4 −2 Col(A) and N ul(A). 7 −4 Final Answer: Any column of A provides a nonzero vector in Col(A) and 1 ∈ N ul(A). 0 Many other answers are possible as well. · ¸ · ¸ 1 3 5 0 1 0 −7 6 Work: We have A = ∼ , so N ul(A) is described by 0 1 4 −2 0 1 4 −2 7 −6 −4 2 x = x3 1 + x4 0 0 1 3

where x3 and x4 are free variables. Putting, say, x3 = 1 and x4 = 0 gives us a vector in N ul(A). −8 −2 −9 2 4 8 and w = 1 . Determine if w ∈ 4.2.24: Problem Restatement: Let A = 6 4 0 4 −2 Col(A). Is w ∈ N ul(A)? Final Answer: w ∈ Col(A) and w ∈ N ul(A). −8 −2 −9 2 2 6 4 8 1 Work: We have [A|w] = ∼ 0 4 0 4 −2 0 say, then we have w = Ax, so w ∈ Col(A). Also, w ∈ N ul(A).

0 2 0 an

1 2 −1 −2 1 2 . Therefore, if x = 1 , 0 0 0 easy computation shows Aw = 0 so

4.2.28: Problem Restatement: Consider the following systems of equations 5x1 + x2 − 3x3 = 0 −9x1 + 2x2 + 5x3 = 1 4x1 + x2 − 6x3 = 9

5x1 + x2 − 3x3 = 0 −9x1 + 2x2 + 5x3 = 5 4x1 + x2 − 6x3 = 45

Assume the first system has a solution. Without using row operations argue that the second system has a solution. Final Answer: Each system has thesame coefficient matrix. Let this matrix be A. Since the 0 first system has a solution, 1 ∈ Col(A). Col(A) is a subspace of R3 , by Theorem 3 9 0 0 0 0 of section 4.2. Therefore, 5 1 ∈ Col(A). But 5 1 = 5 , so 5 ∈ Col(A). 9 9 45 45 Therefore, the second system of linear equations has a solution. Work: None required.

4

Section 4.3 Linearly Independent Sets; Basis 2 1 −7 4.3.4: Problem Restatement: Determine if { −2 , −3 , 5 } a basis of R3 . If it is 1 2 4 not a basis, determine if it is linearly independent and if it spans R3 . Final Answer: The set is a basis of R3 . Work: echelon form (any) matrix to an the corresponding column Row reducing gives us 1 2 4 1 2 4 1 2 4 2 1 −7 −2 −3 −5 ∼ −2 −3 −5 ∼ 0 1 3 ∼ 3 There are 0 1 0 −3 −15 1 2 4 2 1 −7 0 0 2 three pivot columns, so the three vectors are linearly independent. Therefore, they form a basis of R3 . 4.3.12: Problem Restatement: Find a basis for the set of vectors in R2 on the line y = 5x. · ¸ 1 }. Other answers are possible. Final Answer: { 5 · ¸ · ¸ · ¸ x x 1 Work: { : y = 5x with x, y ∈ R} = { : x ∈ R} = {x : x ∈ R} = y 5x 5 · ¸ 1 Span{ }. 5 4.3.14: Problem Restatement: Assuming find A ∼ B, 1 2 −5 11 −3 1 2 4 −5 15 0 2 and B = where A = 1 2 0 0 4 5 3 6 −5 19 −2 0

a basis of Col(A) and a basis of N ul(A) 2 0 4 5 0 5 −7 8 . 0 0 0 −9 0 0 0 0

Final Answer: Since A ∼ B, inspecting B identifies columns 1, 3 and 5 of A as a basis of Col(A). Inspecting echelon form of A computed in the work section below the reduced −4 −2 1 0 7 gives us { 0 , 5 } as a basis of N ul(A) 0 1 0 0 1 2 0 4 0 0 0 1 −7/5 0 . x and x4 are free, so Work: A ∼ B ∼ 0 0 0 0 1 2 0 0 0 0 0 5

x1 = −2x2 − 4x4 , x2 x3 x4 x5

= x2 + 0x4 , = 0x2 + 57 x4 , = 0x2 + x4 , and = 0x2 + 0x4 .

1 0 4.3.16: Problem Restatement: Find a basis for the space spanned by the vectors v1 = 0 , 1 6 5 0 −2 −1 −3 3 1 v2 = −1 , v3 = 2 , v4 = 3 , and v5 = −1 . −1 −4 1 1 Final Answer: The echelon form, computed in the work section below, of A = [v1 v2 v3 4 v5 ] reveals the first three columns provide a basis of Col(A). Thus, {v1 , v2 , v3 } is a basis of Span({v1 , v2 , v3 , v4 , v5 }). Other answers are possible. Work: 1 −2 6 5 0 0 1 −1 −3 3 0 −1 2 3 −1 1 1 −1 −4 1 1 −2 6 5 0 0 1 −1 −3 3 ∼ 0 −1 2 3 −1 0 3 −7 −9 1 1 −2 6 5 0 0 1 −1 −3 3 ∼ 0 0 1 0 2 0 0 −4 0 −8 1 −2 6 5 0 0 1 −1 −3 3 ∼ 0 0 1 0 2 0 0 0 0 0

4.3.32: Problem Restatement: Suppose T is a one to one linear transformation, so that T (x) = T (y) always implies x = y. Show that if the set of images {T (v1 ), T (v2 ), ..., T (vn )} is linearly dependent then {v1 , v2 , ..., vn } is linearly dependent. 6

Final Answer: Assume {T (v1 ), T (v2 ), ..., T (vn )} is linearly dependent, so that there are scalars α1 , α2 , ..., αn , not all zero, such that α1 T (v1 )+α2 T (v2 )+...+αn T (vn ) = 0. Since T is linear, we have α1 T (v1 ) + α2 T (v2 ) + ... + αn T (vn ) = T (α1 v1 + α2 v2 + ... + αn vn ). Also, 0 = T (0) and this gives us T (α1 v1 + α2 v2 + ... + αn vn ) = T (0). Then α1 v1 + α2 v2 + ... + αn vn = 0 because T is one to one. Finally, because not all α1 , α2 , ..., αn are zero, it follows that {v1 , v2 , ..., vn } is linearly dependent and this completes the proof. Work: None required. Section 4.4 Coordinate Systems −4 3 4 −1 4.4.4: Problem Restatement: Find x when B = { 2 , −5 , −7 } and [x]B = 8 . 3 −7 0 2 0 Final Answer: x = 1 . −5 −1 3 4 0 Work: x = PB [x]B = −4 2 + 8 −5 − 7 −7 } = 1 . 0 2 3 −5 · ¸ · ¸ · ¸ 1 5 4 4.4.6: Problem Restatement: Find [x]B when B = { , } and x = . −2 −6 0 · ¸ −6 . Final Answer: [x]B = 2 · ¸ · ¸ · ¸ · ¸ x1 4 1 5 Work: We want [x]B = such that = x1 + x2 . Thus, we solve the x2 0 −2 −6 system

· to get [x]B =

¸ −6 . 2

x1 + 5x2 = 4 −2x1 − 6x2 = 0

3 2 8 4.4.10: Problem Restatement: Let B = { −1 , 0 , −2 }. B is a basis of R3 . Find 4 −5 7 the change-of-coordinate matrix from B to the standard basis of R3 .

7

Final Answer: By definition, the change-of-coordinate matrix from B to the standard basis of 3 2 8 0 −2 . R3 is PB = −1 4 −5 7 Work: None required. 4.4.14: Problem Restatement: The set B = {1 − t2 , t − t2 , 2 − 2t + t2 } is a basis of P2 . Find the coordinate vector of p(t) = 3 + t − 6t2 relative to B. 7 Final Answer: [p(t)]B = −3 . −2 x1 Work: Soln1: Let [p(t)]B = x2 with x1 , x2 , and x3 the required unknown coordinates. x3 We have 3 + t − 6t2 = x1 (1 − t2 ) + x2 (t − t2 ) + x3 (2 − 2t + t2 ) = (x1 + 2x3 ) + (x2 − 2x3 )t + (−x1 − x2 + x3 )t2 . Upon equating constant, linear, and quadratic coefficients, respectively, we are lead to the system of equation x1 + 0x2 + 2x3 = 3 (constant part) 0x1 + x2 − 2x3 = 1 (linear part) −x1 − x2 + x3 = −6 (quadratic part) 7 The solution of this system is [p(t)]B = −3 . −2 Soln2: P2 ' R3 via the standard representation of a quadratic polynomial a0 + a1 t + a0 a2 t2 as the column vector a1 . Under this linear isomorphism B corresponds to a2 1 0 2 3 {v1 , v2 , v3 } = { 0 , 1 , −2 } and p(t) corresponds to w = 1 . Upon −1 −1 1 −6 row reducing the augmented matrix [v1 v2 , v3 | w]to reduced echelon form we get [I3 | 7 [p(t)]B ]. Indeed, from this we get [p(t)]B = −3 . −2 4.4.22: Problem Restatement: Let B = {b1 , b2 , ..., bn } be a basis of Rn . Produce a description of an n × n matrix A that implements the coordinate mapping x 7→ [x]B .

8

Final Answer: Let PB be the usual n × n matrix of column vectors taken from B. Then the coordinate mapping x → [x]B is PB−1 . Work: PB is the change-of-coordinate matrix from B to the standard basis of Rn . We have x = PB [x]B , so [x]B = PB−1 x.

9