Math 2040: Matrix Theory and Linear Algebra II Solutions to Assignment 1 Section 4.1 Vector Spaces and Subspaces
4.1.2. Problem Restatement: Let W W be the union of the first and third quadrants in the xyxy-
plane. That is W is W = {
x
: xy ≥ 0}. y a. If u is in W in W and c and c is any scalar, is c is c u in W ? W ? Why? b. Find specific vectors u and v in W such in W .. This is enough to show W such that u + v is not in W that W that W is is not a vector space. Final Answer: a. cu is in W in W for for all u in W and scalars c.. Proof in work below. W and scalars c b. Let u
0 −1 = and = , say, but there are many other possibilities. 1 0 x cx v
so cu ∈ W just ∈ W and c ∈ R. cu = W just in case cxcy ≥ 0. But y cy Also cxcy = c 2 xy. xy ≥ 0 since u ∈ W . W . Also cxcy xy . Since c2 ≥ 0, it follows that c2xy ≥ 0. Therefore, cxcy ≥ 0 as required; that is, c u ∈ W . W . 0 0 −1 −1 −1 b. and are each in W , + = is not in W . W , but W . 1 0 1 0 1
Work: a. Let u =
4.1.6. Problem Restatement: Determine if the set { p( + t 2 , a ∈ R} is a subspace of p(t)| p( p(t) = a + t P2 . Justify. Final Answer: The set { p( P 2 since the zero vector p(t)| p( p(t) = a + t2 , a ∈ R} is not a subspace of P 2 0 = 0 + 0t 0t + 0t 0t is not in the set. Work: None required.
2t Show the set H of H of all vectors of the form 0 is a subspace −t by finding a spanning set. 2 so H is is a subspace by Theorem 1. H = Span{ 0 }, so H
4.1.10. Problem Restatement:
of R3
Final Answer:
−1
Work: None required.
1
4.1.16. Problem Restatement:
−a + 1 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 . 0 −a + 1 0 . Then a = 1, so a − 6b = 0 implies b = 61 whereas Work: Suppose a − 6b = 2b + a 0 2b + a = 0 implies a contradictory result that b = − 12 . Other counter examples are possible.
4.1.32. Problem Restatement: Let H and K be subspaces of a vector space V . The intersection
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. 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
Work: 1. (Zero) Since 0 ∈ H and 0 ∈ K it follows that 0 ∈ H K . 2. (Addition) Let u, v ∈ H K . Then u, v ∈ H so u + v ∈ H since H is closed under addition. Also, u, v ∈ K so u + v ∈ K since K is closed under addition. Therefore, u + v ∈ H K . Therefore, H K is closed under addition. 3. (Scalar multiplication) Let u ∈ H K and let c ∈ R. u ∈ H , so 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. 1 0 4. (Union example) Let W = Span{ } Span{ }. W is not closed under addi0 1 1 0 1 0 1 tion since
0
∈ W and
1
∈ W , but
2
0
+
1
=
1
∈ / W .
Section 4.2 Null, Column Spaces and Linear Transformations
4.2.2: Problem Restatement:
5 5 Determine if w = −3 is in N ul(A), where A = 13
21 19 23 2 8 14 1
2
Final Answer: w ∈ Nul(A) Work: Aw
25 − 63 + 38 0 = 65 − 69 + 4 = 0 . 40 − 42 + 2 0 b − 5d 2b : b, d ∈ Let W = { 2d + 1
4.2.12: Problem Restatement:
.
R}. Either use an appropriate
d theorem to show W is a vector space, or find a specific example to the contrary. Final Answer: W is not a vector space since it does not contain 0. b − 5d 0 2b Work: If = 0 , then d = 0 from the forth coordinate. But upon substituting 2d + 1 0 d this value for d into the third coordinate, we get 1 = 0, a contradiction.
4.2.22: Problem Restatement: Let A =
1
3 5 0 . Find a nonzero vector in each of 0 1 4 −2
Col(A) and Nul(A).
7 −4 ∈ N ul(A). Any column of A provides a nonzero vector in C ol(A) and 1 0 Many other answers are possible as well. 1 3 5 0 1 0 −7 6 We have A = , so N ul(A) is described by ∼ 0 1 4 −2 0 1 4 −2 7 −6 −4 + x 2 x = x 1 0
Final Answer:
Work:
3
4
0
3
1
where x3 and x4 are free variables. Putting, say, x3 = 1 and x4 = 0 gives us a vector in Nul(A). 4.2.24: Problem Restatement:
−8 Let A = 6
−2 −9 4 8 4 0 4
Col(A). Is w ∈ Nul(A)? Final Answer: w ∈ Col(A) and w ∈ Nul(A).
−8 We have [A|w] = 6
2 2 −2 −9 4 8 1 ∼ 0 Work: 4 0 4 −2 0 say, then we have w = Ax, so w ∈ Col(A). Also, w ∈ Nul(A).
2 and w = 1 . Determine if w ∈ −2
0 2 0 an
2 −1 − 12 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 the same 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 5 , so 5 ∈ Col(A). of section 4.2. Therefore, 5 1 ∈ Col(A). But 5 1 = 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
4.3.4: Problem Restatement:
2 1 −7 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: Row reducing the corresponding column matrix to an (any) echelon form gives us 1 2 4 2 1 −7 1 2 4 1 2 4 1 3 ∼ −2 −3 −5 ∼ −2 −3 −5 ∼ 0 0 1 3 There are 1 2 4 2 1 −7 0 −3 −15 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
Final Answer: { Work:
5
}. Other answers are possible.
x : y = 5x with x, y ∈ { y 1
Span{
5
x
R} = {
5x
: x ∈ R} = {x
1 5
: x ∈ R} =
}.
4.3.14: Problem Restatement: Assuming A ∼ B , find a basis of Col(A) and a basis of N ul(A)
1 2 where A = 1
2 −5 11 −3 1 2 0 4 5 4 −5 15 2 0 0 5 −7 8 and B = . 2 0 4 5 0 0 0 0 −9 3 6 −5 19 −2 0 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 the reduced echelon form of A computed in the work section below −2 −4 1 0 7 0 , gives us { } as a basis of N ul(A) 5 0 1 0 0
Work:
1 0 A ∼ B ∼ 0
2 0 0 0 0
0 4 0 1 −7/5 0 0 0 1 0 0 0
. x and x are free, so 2
4
5
x1 = −2x2 − 4x4 , x2 = x 2 + 0x4 , x3 = 0x2 + 57 x4, x4 = 0x2 + x4 , and x5 = 0x2 + 0x4 . 4.3.16: Problem Restatement: Find a basis for the space spanned by the vectors v1
1 0 , = 0 1
−2 6 5 0 1 , v = −1 , v = −3 , and v = 3 . v = −1 2 3 −1 2
3
4
5
1 −1 −4 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. 1
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 01 −23 −76 −95 10 0 1 −1 −3 3 ∼ 0 0 1 0 2 01 −0 2 −4 6 0 5 −08 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 α 1 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 α 1 v1 + α2 v2 + ... + αn vn = 0 because T is one to one. Finally, because not all α 1 , 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
−1 3 4 −4 Find x when B = { 2 , −5 , −7 } and [x] = 8 . 0 2 3 −7 0 x = 1 . −5 −1 3 4 0 x = P [x] = −4 2 + 8 −5 − 7 −7 } = 1 . 0 2 3 −5 1 5 4 Find [x] when B = { . , } and x = 0 −2 −6 −6 [x] = . 2 x 4 1 5
4.4.4: Problem Restatement:
B
Final Answer:
Work:
B
B
4.4.6: Problem Restatement: Final Answer:
B
B
1
Work: We want [x]B =
x2
such that
0
= x 1
−2
+ x2
−6
. Thus, we solve the
system
x1 + 5x2 = 4 −2x1 − 6x2 = 0 to get [x]B =
−6 2
.
4.4.10: Problem Restatement:
3 2 8 Let B = { −1 , 0 , −2 }. B is a basis of
4 7 −5 the change-of-coordinate matrix from B to the standard basis of R3 .
7
R3 . Find
Final Answer: By definition, the change-of-coordinate matrix from B to the standard basis of
3 = −1
2 8 0 −2 R3 is P B 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 [ p(t)] = −3 . −2 x Let [ p(t)] = x with x , x , and x the required unknown coordinates.
Final Answer:
B
1
Work: Soln1:
B
2
1
2
3
x3 We have 3 + t − 6t2 = x1(1 − t2 ) + x 2 (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: P 2 R3 via the standard representation of a quadratic polynomial a0 + a 1t + a0 2 a1 . Under this linear isomorphism B corresponds to a2 t as the column vector a2 1 0 2 3 0 , 1 , −2 } and p(t) corresponds to w = 1 . Upon {v1 , v2 , v3} = { 1 −1 −1 −6 row reducing the augmented matrix [v1 v2 , v 3 | w] to reduced echelon form we get [ I 3 | 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 → [x]B .
8
Final Answer: Let P B be the usual n × n matrix of column vectors taken from B. Then the
coordinate mapping x → [x]B is P B−1 . Work: P B is the change-of-coordinate matrix from B to the standard basis of Rn . We have x = P B [x]B , so [x]B = P B−1 x.
9
