MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 1. HW 1: Due September 4 1.1.21. Suppose v, w R n and c is a scalar. Prove that Span(v + cw, w) = Span(v, w). We must prove two things: that every element of Span(v + cw, w) is in Span(v, w), and that every element of Span(v, w) is in Span(v, w). If x Span(v + cw, w), then x = a(v + cw) + bw for some scalars a, b R. Then x = av + (b + ac)w, so x is in Span(v, w). Similarly, if x Span(v, w), then x = dv+ew for some scalars d, e R. We can rearrange this as x = dv + dcw dcw + ew = d(v + cw) + (e dc)w. Hence, x is in Span(v + cw, w). 1.1.29. (a) Using only the properties listed in Exercise 28, prove that for any x R n, we have 0x = 0. We will be extremely formalistic in this exercise. On most problems, you don t have to show quite so much work, but the idea here is to see carefully that the properties in Exercise 28 are really all that s required to do linear algebra. For any vector x R n, we have: 0x + x = 0x + x by (h) = (0 + 1)x by (g) = 1x = x by (g) again Now add x (which exists by property (d)) to both sides of the equation: (0x + x) + ( x) = x + ( x) 0x + (x + ( x)) = x + ( x) 0x + 0 = 0 x = 0 by (b) by (d) by (a) and (c) (b) Prove that ( 1)x = x. First, we have to observe that x is uniquely characterized by the property that x+( x) = 0. If y were some other vector with the property that x + y = 0, then we could write: y + (x + ( x)) = (y + x) + ( x) y + 0 = 0 + ( x) y = xby (c). 1 by (b) by (d) and our assumption on y

2 MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS Hence, it suffices to show that x + ( 1)x = 0. This is seen as follows: as required. x + ( 1)x = 1x + ( 1)x by (h) = (1 1)x by (g) = 0x = 0 by part (a) of the problem 1.2.16. Let y R n. If x y = 0 for all x R n, then prove that y = 0. Let s follow the hint and consider the dot products of y with some strategically chosen vectors. For i = 1,..., n, let e i be the vector with a 1 in the i th entry and 0 everywhere else. Observe that e i y = 0y 1 + + 0y i 1 + 1y i + 0y i+1 + + 0y n = y i. Thus, since e i y = 0 for all i (which is true by assumption), we see that y 1 = = y n = 0, and hence y = 0. 2. HW 2: Due September 11 1.3.12. Suppose a 0 and P R 3 is the plane through the origin with normal vector a. Suppose P is spanned by u and v, and assume that u v = 0. (a) Show that for every x P, we have x = proj u (x) + proj v (x). Since x P, and P = Span(u, v), we can write x = su + tv for some scalars s, t. We will try to figure out what s and t must be. Compute the dot product of x with u: x u = (su + tv) u = s(u u) + t(v u) = s(u u) so s = x u x v. A similar computation shows that t =. Plugging in these values for s and t, u u v v we have: x = x u u u u + x v v v v = proj u (x) + proj v (x) by definition. (b) Show that for any x R n, we have x = proj a (x) + proj u (x) + proj v (x). Following the hint, let w = x proj a (x), which is just the perpendicular part of x with respect to a. Thus, w a = 0, so w P. Hence, by applying the previous part, we can write and therefore w = proj u (w) + proj v (w), x = proj a (x) + proj u (w) + proj v (w). We just have to check that proj u (x) = proj u (w) and proj v (x) = proj v (w), and then we ll be done.

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 3 For the first of these equations, let s write proj a (x) = ca, where c = x a. We have: a a proj u (x) = x u u u u (w + ca) u = u u u = w u u u u + ca u u u u = w u u u u = proj u (w) as required. Here we used the fact that a u = 0, which is true since u P. A similar reasoing applies to show that proj v (x) = proj v (w). 1.4.15. (a) Prove or give a counterexample: If A is an m n matrix and x R n satisfies Ax = 0, then either every entry of A is zero or x = 0. This is definitely false: it would be saying that no homogeneous linear equation can have any nonzero solutions! As a very basic counterexample, let A = [ 1 1 ] [ 1, and x =, so 1] that Ax = [1 1] = [0] (which we re considering as a vector in R 1 ). (b) Prove or give a counterexample: If A is an m n matrix, and Ax = 0 for every vector x R n, then every entry of A is 0. Following the hint, notice that the entries of Ax are the dot products A i x, where A i are the rows of A. If Ax = 0 for all x, then A i x = 0 for all x, and therefore A i = 0 by Problem 1.2.16 (from last week, solved above). Hence we deduce that A is the zero matrix. 3. HW 3: Due September 18 1.5.12. In each case, give positive integers m and n and an example of m n matrix A with the stated property, or explain why none can exist. (a) Ax = b is inconsistent for every b R m. If b = 0, then the solution Ax = b always has at least one solution, namely x = 0. Therefore, this can t happen. (b) Ax = b has one solution for every b R m. This will be true for any nonsingular n n matrix. The most basic example is m = n = 1 and A = [1]. (c) Ax = b has no solutions for some b R m and one solution for every other b R m. This definitely can t happen. For instance, if Ax = b has no solutions, then Ax = 2b also has no solutions, since if x were a solution to Ax = 2b, then 1 x would be a solution to 2 Ax = b. (d) Ax = b has infinitely many solutions for every b R m. This will be true for any m n matrix A with m < n and rank(a) = m. An example is m = 1, n = 2, A = [ 1 1 ]. (e) Ax = b is inconsistent for some b R m and has infnitely many solutions whenever it is consistent.

4 MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS This will be true for any m n matrix A where rank(a) < m (which guarantees that it s sometimes inconsistent) and rank(a) < n (which guarantees[ that] there are infinitely many 1 0 solutions). For instance, we could take m = n = 2 and A =. 0 0 (f) There are vectors b 1, b 2, b 3 R m such that Ax = b 1 has no solutions, Ax = b 2 has one solution, and Ax = b 3 has infnitely many solutions. We saw in class that this cannot happen. 1.5.13. Suppose A is an m n matrix with rank m, and v 1,..., v k R n with Span(v 1,..., v k ) = R n. Prove that Span(Av 1,..., Av k ) = R m. We need to show that every vector in R n can be written as a linear combination of Av 1,..., Av k. Since rank(a) = m, for any b R n, we know that the system Ax = b has a solution, which means that b = Aw for some vector w R n. Since the vectors v 1,..., v k span all of R n, we can write w = c 1 v 1 + + c k v k for some scalars c 1,..., c k. We then observe: b = Aw = A(c 1 v 1 + + c k v k ) = c 1 Av 1 + + c k Av k. So b Span(Av 1,..., Av k ), as required. 1.5.14. Let A be an m n matrix with row vectors A 1,..., A m. (a) Suppose A 1 + +A m = 0. Deduce that rank(a) < m. First proof: For any vector x = (x 1,..., x n ), we have 0 = (A 1 + + A m ) x = A 1 x + + A m x, which is the sum of the entries of Ax. This means that if Ax = b has solutions, then the sum of the entries in b must be zero. Equivalently, if b is any vector whose sum of entries is nonzero, then Ax = b has no solutions. This implies that rank(a) < m. Second proof: Let s perform some row operations to A. First, add a multiple of each of the first m 1 rows to the last row, and call the resulting matrix B. By assumption, the new m th row will be all zeros. If we then perform Gaussian elimination to obtain any echelon form of B (which is an echelon form of A), there will be at least one row of zeros at the bottom, and therefore rank(a) < m. (b) More generally suppose there is some linear combination c 1 A 1 + +c m A m = 0, where some c i 0. Show that rank(a) < m. We can easily adapt the first proof from above to show that if Ax = b has solutions, then c 1 b 1 + + c m b m = 0. If we choose b = (0,..., 1,..., 0), where the 1 is in the i th entry, then we see there are no solutions. 4. HW 4: Due September 27 [ ] a b 2.1.7. Find all 2 2 matrices A = satisfying: c d (a) A 2 = I

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 5 We need to solve four non-linear equations: a 2 + bc = 1 ab + bd = 0 ac + cd = 0 bc + d 2 = 1. Combining the first and fourth equations tells you that a 2 = d 2 = 1 bc, which means that a = ±d. We can consider two cases: either a = d 0, or a = d. In the first case, the second and third equations tell us that b = c = 0, and therefore the [ first and ] fourth [ equations ] say that a 2 = d 2 = 1. Hence, the only two matrices we obtain are 1 0 1 0 and. 0 1 0 1 In the second case, the second and third equations don t give any constraint on b and c. The only constraint is that 1 bc 0, since otherwise a 2 and d[ 2 would be negative. For ] any ± 1 bc b b and c with bc 1, we obtain two possible matrices, namely c. 1 bc (b) A 2 = O We proceed similarly to the preceding problem, where now the right-hand sides of all four equations are 0. As before, we deduce that a 2 = d 2. If a = d, then the second and third equations give b = c = 0, and then the first and fourth give a = d = 0. If a = d, then b and [ c can be any numbers ] with bc 0 (i.e. they have opposite signs), and then we get ± bc b A = c. bc (c) A 2 = I 2 In this case, note that we [ can t have a = d 0, since ] that would force a 2 = d 2 = 1. ± 1 bc b Hence the only solutions are c. 1 bc 4.1. 2.1.14. Find [ all 2 ] 2 matrices A that commute [ with ] all 2 2 matrices B. a b e f Suppose A =, and that for every B =, we have AB = BA. This means c d g h that for all possible e, f, g, h R, we have: ae + bg = ae + cf ce + dg = ag + eh af + bh = be + df cf + dh = bg + dh. In particular, we can plug in (e, f, g, h) = (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), or (0, 0, 0, 1). This tells us that b = c = 0 and a = d, i.e. A must be a multiple of the identity matrix. (And we already know that any multiple of the diagonal matrix commutes with all matrices B.) 2.2.7. (a) Calculate A θ A φ and A φ A θ. [ ] [ ] cos θ sin θ cos φ sin φ A θ A φ = sin θ cos θ sin φ cos φ [ ] cos θ cos φ sin θ sin φ sin θ cos φ cos θ sin φ = sin θ cos φ + cos θ sin φ cos θ cos φ + sin θ sin φ

6 MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS And A φ A θ equals the same matrix, as can be seen by swapping the roles of θ and φ everywhere. (b) Use your answer to part (a) to derive the addition formulas for sine and cosine. Geometrically, rotating the plane by θ and then rotating it by φ is the same as rotating it by θ + φ all at once. Hence, the above matrix is equal to A θ+φ. This means, in particular, that cos(θ + φ) = cos θ cos φ sin θ sin φ sin(θ + φ) = sin θ cos φ + cos θ sin φ. (Note: If, like me, you always have difficulty remembering the angle addition formulas, you can easily remember them using this method!) 2.2.8. For 0 θ π, prove that A θ x = x and that the angle between x and A θ x equals θ. To avoid writing lots of square roots, let s just compute A θ x 2. We have: [ ] A θ x 2 = x1 cos θ x 2 sin θ 2 x 1 sin θ + x 2 cos θ = (x 1 cos θ x 2 sin θ) 2 + (x 1 sin θ + x 2 cos θ) 2 = x 2 1 cos 2 θ 2x 1 x 2 cos θ sin θ + x 2 2 sin 2 θ x 2 1 sin 2 θ + 2x 1 x 2 cos θ sin θ + x 2 2 cos 2 θ = (x 2 1 + x 2 2)(cos 2 θ + sin 2 θ) = x 2 1 + x 2 2 = x 2. If φ denote the angle between x and A θ x, then cos φ = x A θx x A θ x = x 1(x 1 cos θ x 2 sin θ) + x 2 (x 1 sin θ + x 2 cos θ) x 2 = (x2 1 + x 2 2) cos θ x 2 = cos θ Since θ and φ are both between 0 and π and both have the same cosine, we must have θ = φ. 2.3.16. Suppose A is an n n matrix satisfying A 10 = O. Prove that the matrix I n A is invertible.

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 7 There s actually nothing special about 10 here; let s assume A m = O for some integer m > 1. The trick is first to remember some facts about factoring polynomials: and in general t 2 1 = (t 1)(t + 1) t 3 1 = (t 1)(t 2 + t + 1) Analogous formulas hold for matrices: t 4 1 = (t 1)(t 3 + t 2 + t + 1) t m 1 = (t 1)(t m 1 + t m 2 + + t + 1). A m I n = (A I n )(A m 1 + A m 2 + + A + I n ). We can prove this by just expanding out the right side and canceling a lot of terms. Now if A m = O, we see that and hence I n = (A I n )(A m 1 + A m 2 + + A + I n ) I n = (I n A)(A m 1 + A m 2 + + A + I n ) (I n A) 1 = A m 1 + A m 2 + + A + I n. 5. HW 5: Due October 11 2.4.12. Suppose A and B are two m n matrices with the same reduced echelon form. Show that there exists an invertible matrix E so that EA = B. Is the converse true? Let C denote the reduced echelon form of both A and B. Since row operations can be realized by multiplying on the left by elementary matrices, there are elementary matrices E 1,..., E k such that C = E k E 1 A and elementary matrices E 1,..., E l such that C = E l E 1B. Equating these two, and using the fact that elementary matrices are invertible, we have: E k E 1 A = E l E 1B A = (E 1 ) 1 (E k ) 1 E l E 1B Hence we define E = (E 1 ) 1 (E k ) 1 E l E 1, which is invertible since it is the product of invertible matrices. To see the converse, suppose EA = B, where E is invertible. This means that E can be row reduced to the identity. It follows that E is equal to a product of elementary matrices. Thus, there is a sequence of row operations taking A to B. By the uniqueness of reduced echelon forms, we see that A and B must have the same reduced echelon form. 2.5.12. Suppose A is a symmetric n n matrix. If x, y R n are vectors satisfying Ax = 2x and Ay = 3y, show that x and y are orthogonal. Using the fact that A is symmetric, we have x Ay = x T Ay = y T A T x = y T Ax = Ax y. Also, x Ay = x 3y = 3(x y), and Ax y = 2x y = 2(x y). Putting these together, we see that 2(x y) = 3(x y), and hence x y = 0.

8 MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 6. HW 6: Due October 16 3.1.13. Let V R n be a subspace. Show that V (V ). Do you think more is true? For any vector v V, and any vector w V, we have v w = 0 (since w is orthogonal to every vector in V ). That is, v is orthogonal to every vector in V, so v (V ). This shows that V (V ). The reverse inclusion was shown in class. 3.1.14. Let V and W be subspaces of R n with the property that V W. Prove that W V. For any vector u W, we have u w = 0 for every vector w W. In particular, this is true for every w V. Thus, u V. 3.2.10. Let A be an m n matrix, and let B be an n p matrix. (a) Prove that N(B) N(AB). For any vector v N(B), we have Bv = 0, and hence AB(v) = A(Bv) = 0. (Note: here the 0 in the first equation is in R n, and the 0 in the second is in R m.) Hence, v N(AB). (b) Prove that C(AB) C(A). If v C(AB), then there is a vector w R p such that (AB)w = v. (I.e., the system (AB)x = v is consistent.) Rewriting this, we see that A(Bw) = v, which means that the system Ax = v is consistent as well. Hence, v C(A). (c) If A is n n and nonsingular, prove that N(B) = N(AB). We already know from part (a) that N(B) N(AB); we need to prove that N(AB) N(A). For any v N(AB), we have ABv = 0. Multiplying both sides on the left by A 1 (which exists because A is nonsingular), we see that Bv = 0, so v N(B). (d) If B is n n and nonsingular, prove that C(AB) = C(A). We already know from part (b) that C(AB) C(A); we just need to show that C(A) C(AB). For any vector v C(A), there is a vector w R n such that Aw = v. Then AB(B 1 w) = Aw = v, so v C(AB) as well. 3.2.11. Let A be an m n matrix. Prove that N(A T A) = N(A). The inclusion N(A) N(A T A) follows from the previous problem; we need to see that N(A T A) N(A). If v N(A T A), then A T Av = 0, so Av N(A T ). Also, Av C(A), by definition. And since C(A) = N(A T ), we deduce that Av = 0, so v N(A), as required. 7. HW 7: Due October 23 3.3.10. Suppose v 1,..., v k are nonzero vectors such that v i v j = 0 whenever i j. Prove that {v 1,..., v k } is linearly independent. Suppose that c 1 v 1 + + c k v k = 0; we will show that this implies that c 1 = = c n = 0, which means that the vectors v 1,..., v k are linearly independent. For each i = 1,..., n, taking the dot product of this equation with v i gives: c 1 v 1 v i + + c i v i v i + + c k v k = 0. By hypothesis, all of the terms in this sum except for the ith are 0, so c i (v i v i ) = 0. Moreover, since v i 0, v i v i 0, so we can divide and deduce c i = 0.

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 9 3.3.11. Suppose v 1,..., v n are nonzero, mutually orthogonal vectors in R n. (a) Prove that they form a basis for R n. By the previous problem, we see that v 1,..., v n are linearly independent, and any n linearly independent vectors in R n must span R n. (b) Given any x R n, give an explicit formula for the coordinates of x with respect to the basis {v 1,..., v n }. Suppose x = a 1 v 1 + + a n v n. By definition, the coordinates of x are the coefficients a 1,..., a n. Just as in the previous problem, for each i = 1,..., n, we take the dot product of each side with v i : x v i = a 1 v 1 v i + + a i v i v i + + a n v n v i = a i v i v i and therefore a i = x v i v i v i. (c) Deduce from your answer to part (b) that x = n i=1 proj v i (x). We have n n x v i n x = a i v i = v i = proj v i v vi (x). i i=1 i=1 3.3.15. Suppose k > n. Prove that any k vectors in R n must form a linearly dependent set. Let v 1,..., v k be the vectors, and let A be the n k matrix whose columns are v 1,..., v k. Since rank(a) n < k, there must be a nonzero vector c with Ac = 0. But this means that c 1 v 1 + + c k v k = 0 and the coefficients c i are not all zero, as required. 3.3.19. Let A be an n n matrix. Prove that if A is nonsingular and {v 1,..., v k } is linearly independent, then {Av 1,..., Av k } is likewise linearly independent. Give an example to show that the result is false if A is singular. Suppose we have a linear relation c 1 (Av 1 ) + + c k (Av k ) = 0. We may rewrite this as A(c 1 v 1 + + c k v k ) = 0. Since A is nonsingular, the only solution to Ax = 0 is the zero vector, so c 1 v 1 + + c k v k = 0. Therefore, c 1 = = c k since v 1,..., v k are linearly independent. [ We ] have thus[ shown that Av 1,..., Av k are linearly independent as well. 1 0 0 If A = and v 0 0 1 =, then Av 1] 1 = {0 0}, so the set {Av 1 } is linearly dependent even though {v 1 } is linearly independent. Note: The proof of 3.3.21 is essentially the same. 3.3.22. Let A be an n n matrix, and suppose v 1, v 2, v 3 R n are nonzero vectors such that Av 1 = v 1, Av 2 = 2v 2, and Av 3 = 3v 3. Prove that {v 1, v 2, v 3 } is linearly independent. Following the hint, we ll first show that {v 1, v 2 } is linearly independent. If not, then v 2 is a nonzero multiple of v 1, say v 2 = av 1, where a 0. On the one hand, we have Av 2 = 2v 2 = 2av 1, but on the other hand Av 2 = A(av 1 ) = av 1. Since a 2a and v 1 0, this is impossible. Hence, {v 1, v 2 } is linearly independent. i=1

10 MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS Next, if {v 1, v 2, v 3 } is linearly dependent, then v 3 must be in the span of v 1, v 2 : say v 3 = bv 1 + cv 2, where b and c are not both zero. Then Av 3 = 3v 3 = 3bv 1 + 3cv 2, but at the same time Av 3 = A(bv 1 + cv 2 ) = bv 1 + 2cv 2. Combining these: and hence 3bv 1 + 3cv 2 = bv 1 + 2cv 2 2bv 1 + cv 2 = 0. Since {v 1, v 2 } is linearly independent, this forces b = c = 0, a contradiction. {v 1, v 2, v 3 } is linearly independent. Hence, 3.4.20. Let U and V be subspaces of R n. Prove that if U V = {0}, then dim(u + V ) = dim U + dim V. Suppose that dim U = k and dim V = l; we will prove that dim(u + V ) = k + l. Let {u 1,..., u k } be a basis for U and {v 1,..., v l } be a basis for V. We claim that {u 1,..., u k, v 1,..., v l } is a basis for U + V. It is simple to check that these vectors span U +V. Any vector x U +V can be written as u + v, where u U and v V. We can write u = a 1 u 1 + + a k u k and v = b 1 v 1 + + b l v l for some a 1,..., a k, b 1,..., b l R, and therefore x = u = a 1 u 1 + + a k u k + b 1 v 1 + + b l v l as required. To see that {u 1,..., u k, v 1,..., v l } is linearly independent, suppose we have a linear relation: c 1 u 1 + + c k u k + d 1 v 1 + + d l v l = 0. Rewrite this as c 1 u 1 + + c k u k = d 1 v 1 + d l v l. The left-hand side is an element of U, and the right-hand side is an element of V, and they are equal to each other. Since U V = {0}, each side must equal 0. Since {u 1,..., u k } and {v 1,..., v l } are each linearly independent, we see that c 1 = = c k = d 1 = = d l = 0, as required.