Mathematics 206 Solutions for HWK 13b Section 5.2

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1 Mathematics 206 Solutions for HWK 13b Section 5.2 Section Problem 7ac. Which of the following are linear combinations of u = (0, 2,2) and v = (1, 3, 1)? (a) (2, 2,2) (c) (0,4, 5) Solution. Solution by inspection doesn t look feasible to me, so solve two systems of equations. We may as well solve them both at the same time. We want to know whether there exist scalars a 1, a 2 for which a 1 (0, 2, 2) + a 2 (1,3, 1) = (2, 2,2) and whether there exist scalars c 1, c 2 for which c 1 (0, 2, 2) + c 2 (1,3, 1) = (0, 4, 5). When these equations are rewritten as systems of equations, the two coefficient matrices are identical. Combining the two augmented matrices into a single matrix, we have the doubly augmented matrix The reduced row-echelon form for this matrix is From the bottom row and the last column we see that it s impossible to find the desired c 1, c 2, so the vector in part (c) is not a LC of u and v. However, the vector in (a) is a LC of u and v. From the first three columns we see that values for a 1 and a 2 = 2 should be be 2. In other words (2,2,2) = 2u + 2v. This result is easy to check. Section Problem 8ac. Express the following as linear combinations of u = (2,1, 4), v = (1, 1,3), and w = (3, 2,5). (a) ( 9, 7, 15) (c) (0,0, 0) Solution. Part (c) can be solved by inspection: one solution is 0 = 0u + 0v + 0w. Page 1 of 7 A. Sontag April 6, 2002

2 For part (a) we need to find scalars k 1, k 2, k 3 for which k 1 (2, 1, 4) + k 2 (1, 1, 3) + k 3 (3,2, 5) = ( 9, 7, 15). Rewrite this as a system of equations with augmented matrix The reduced row-echelon form is This gives k 1 = 2, k 2 = 1, and k 3 = 2 and ( 9, 7, 15) = 2u + v 2w. Note that this result is easy to check. Section Problem 9ac. Express the following as linear combinations of p 1, p 2, and p 3,given that p 1 (x) = 2 + x + 4x 2, p 2 (x) = 1 x + 3x 2, and p 3 (x) = 3 + 2x + 5x 2. (a) q, with q(x) = 9 7x 15x 2 (c) 0 (in other words the constant function 0) Solution. Again, part (c) is easily solved by inspection: 0 = 0p 1 + 0p 2 + 0p 3. For part (a), we need to find scalars k 1, k 2, k 3 such that, for every real number x, k 1 (2 + x + 4x 2 ) + k 2 (1 x + 3x 2 ) + k 3 (3 + 2x + 5x 2 ) = 9 7x 15x 2 Combining terms and equating the coefficients of like powers of x we find k k 2 = k 3 15 Notice that the augmented matrix is the same as that for 8(c). This gives us the single solution k 1 = 2, k 2 = 1, and k 3 = 2. In other words, q = 2p 1 + p 2 2p 3. Again, it s easy to check this result. Page 2 of 7 A. Sontag April 6, 2002

3 Section Problem 10ab. Which of the following are linear combinations of A =, B =, C = (a) M = 1 8 (b) Solution. A, B, C. Part (b) is easy: 0 = 0A + 0B + 0C, so (a) is a linear combination of the specficied For part (a), the question asks whether there exist scalars k 1, k 2, k 3 for which k 1 A + k 2 B + k 3 C = M. By equating individual entries, we see that this last equation is equivalent to the system of equations whose augmented matrix is The reduced row-echelon form is Thus M = A+2B 3C, a linear combination of the given three matrices. Again, it s easy to check this result. Section Problem 12. Let f and g be defined by f(x) = cos 2 x and g(x) = sin 2 (x). Which of the following lie in the space spanned by f and g? (a) cos 2x (b) 3 + x 2 (c) 1 (d) sin x (e) 0 Solution. Part (e) is easiest: 0 = 0f + 0g. Part (c) uses the familiar identity sin 2 x + cos 2 x = 1 for every x. Thus 1 = f + g. Similarly part (a) can be solved by using the double-angle formula for the cosine function: cos 2x = cos 2 x sin 2 x for every x, so cos 2x = f g. Page 3 of 7 A. Sontag April 6, 2002

4 The functions in (b) and (d) look like they won t work. For (b), here s one argument we could use. The linear combination af + bg will give a function whose values all lie between a b and a + b, since a cos 2 x + bsin 2 x a + b according to the triangle inequality. The function 3 + x 2, on the other hand will have values that are not bounded above: 3 + x 2 as x ±. So it s impossible to have 3 + x 2 = acos 2 x + bsin 2 x for every x. ( Other arguments are possible. For instance, you could use specific values of x to reach a contradiction.) For part (d) one could argue as follows. Suppose we did have sin x = a cos 2 x + bsin 2 x for every x. Using the value x = 0 we find a = 0 and the assumption reduces to sin x = bsin 2 x for every x. But then the left-hand side will have some values positive and some negative, while the right hand side will either be always 0 or always 0, according to whether b 0 or b 0. That s impossible. (Other arguments are possible.) Final conclusion: the functions in (a), (c), (e) lie in the specified space. Those in (b), (d) do not. Section Problem 15. u = ( 1, 1,1) and v = (3, 4,4). Find an equation for the plane spanned by the vectors Solution. The specified plane is the vector space consisting of all points of the form s( 1,1, 1)+ t(3,4, 4), where s and t are real scalars. This gives a parametric description of the plane, but we want an equation of the form ax + by + cz = 0. Reclaiming our knowledge of Math 205, we recast the problem: find the equation for the plane through the points (0,0, 0),( 1, 1,1), (3,4, 4). Consult the Math 205 text or consult Section 3.5 in our text for ways to find the required equation. One way is as follows. Compute the cross product ( 1, 1,1) (3, 4,4), which works out to be 7j 7k. The vector j k = (0,1, 1) can therefore be used as a normal vector n for the desired plane and (0, 0,0) is a convenient point known to lie on the plane. The equation of the plane is therefore n (x 0, y 0, z 0) = 0 or (0,1, 1) (x, y, z) = 0 or y = z. Section Problem 16. Find parametric equations for the line spanned by the vector u = (3, 2,5). Solution. The line in question is the vector space consisting of all points of the form t(3, 2,5) for t a real scalar. So a set of parametric equations would be x = 3t, y = 2t, z = 5t, < t <. Other correct answers are possible. Page 4 of 7 A. Sontag April 6, 2002

5 Section Problem 19. Use Theorem to show that v 1 = (1, 6,4), v 2 = (2,4, 1), v 3 = ( 1,2, 5) and w 1 = (1, 2, 5), w 2 = (0, 8,9) span the same supspace of R 3. Solution. According to the cited theorem the given two sets of vectors will span the same subspace iff each of the vectors v 1, v 2, v 3 is a linear combination of w 1 and w 2 and each of the vectors w 1, w 2 is a linear combination of v 1, v 2, v 3. By inspection (or if need be, by solving a system of equations), we can see that v 1 = w 1 + w 2 v 2 = 2w 1 + w 2 v 3 = w 1 Also by inspection (or by solving the relevant system of equations) we see that w 1 = v 3 w 2 = v 1 + v 3 According to Theorem 5.2.4, span({v 1, v 2, v 3 }) = span({w 1, w 2 }). In other words every vector that is a linear combination of v 1, v 2, v 3 is also a linear combination of w 1, w 2, and vice versa. Section Problem 21c. Show that the set of all everywhere differentiable functions that satisfy f + 2f = 0 forms a subspace of F(, ). Solution. Let W denote the specified set of functions. In other words, a function f in F(, ) belongs to W iff it is differentiable everywhere and satisfies the condition f + 2f = 0. Step 1. The constant function 0 is differentiable everywhere and its derivative is the constant function 0. So when twice this function is added to the derivative of this function, the net result is still the constant function 0. In other words the constant function does belong to the set W. Step 2. Suppose that f and g both belong to W. Then we know that f and g are each differentiable everywhere. A calculus theorem tells us that if two functions are each differentiable at a point, then their sum is differentiable at that point and the sum of the derivatives is the derivative of the sum. Therefore f + g must be differentiable everywhere and satisfy (f + g) = f + g. From the fact that f and g belong to W we also know that f + 2f = 0 and g + 2g = 0. Therefore Page 5 of 7 A. Sontag April 6, 2002

6 (f + g) + 2(f + g) = f + g + 2f + 2g = f + 2f + g + 2g = = 0. We shown that f + g belongs to W. Since f and g were given arbitrarily, we ve shown that W is closed under addition. Step 3. Suppose that f belongs to W and k is a scalar. Since f belongs to W, f is differentiable everywhere. A theorem from calculus tells us that kf is differentiable everywhere and satisfies (kf) = kf. We also know that f + 2f = 0. It follows that (kf) + 2(kf) = kf + 2kf = k(f + 2f) = k0 = 0. We ve shown that kf belongs to W. Since k and f were given arbitrarily, we ve shown that W is closed under scalar multiplication. Steps 1,2, and 3 show that W is a subspace of F(, ). Section Problem 23ce. counterexample. True or False? Justify answer with an argument or a (c) If S is a finite set of vectors in a vector space V, then span(s) must be closed under addition and scalar multiplication. This is true, according to Theorem and the definition of span(s), which appears on page 217. The theorem tells us that span(s) is a subspace of V, and subspaces are always closed under addition and scalar multiplication. (e) If span(s 1 ) = span(s 2 ), then S 1 = S 2. This is definitely false. The same vector space can have lots of different spanning sets. A very simple example to show this is the following: V = R 2, S 1 = {(1,0), (0,1)}, S 2 = {(1, 0),(0, 2)}. In this example, span(s 1 ) = span(s 2 ) = V yet S 1 S 2. Lots of other examples, many of them more interesting than this one, are possible. Section Problem 24. (a) Under what conditions will two vectors in R 3 span a plane? A line? Two vectors in R 3 will span a plane iff neither is a linear combination of the other. (For then neither can be discarded without diminishing the span.) They span a line iff at least one of the two is nonzero and at least one is a linear combination of the other. (For then one can be discarded without diminishing the span. If one vector is the zero vector, discard it. Otherwise, each will be a linear combination of the other and one can discard either one. The remaining vector will have the same span as the original two vectors had. The span of a single nonzero vector is a line.) Page 6 of 7 A. Sontag April 6, 2002

7 (b) Under what conditions will it be true that span{u} = span{v}? Either both vectors must be the zero vector, in which case the span is just {0}, or one must be a nonzero scalar multiple of the other, which will automatically make each a nonzero scalar multiple of the other. With u = kv and k 0, each vector au could be rewritten as akv while each bv could be rewritten as b ku. A compressed version of what s needed is that each of the vectors must be a scalar multiple of the other, the answer that s given in the text. (c) If Ax = b is a consistent system of m equations in n unknowns, under what conditions will it be true that the solution set is a subspace of R n? Explain. Given that Ax = b is consistent, we know that the solution set is a nonempty subset of R n. If b = 0, the system is homogeneous and Theorem tells us that the solution set is a subspace of r n. Ifb 0 it is pretty easy to check that the solution set can t be closed under either addition or scalar multiplication. So the solution set is a subspace of R n iff b = 0. Note that in this case the given system is automatically consistent. Alternate reasoning, with the same result, can be found in the text. Section Problem 26. (a) Let M 22 be the vector space of 2 2 matrices. Find four matrices that span M 22. Since a b = a c d + b c + d 0 1 we can use the four matrices, 0 1,,. 0 1 Many other correct answers are possible. (b) In words, describe a set of matrices that spans M nn. The set consisting of all n n matrices, each of which has exactly one of its entries equal to 1 and all of the remaining entries equal to 0. Page 7 of 7 A. Sontag April 6, 2002

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