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Unformatted text preview: ma253, Fall 2010 — Problem Set 1 Solutions
1. We work with the linear space R2 . Fix two numbers a and b and consider the function f : R2 −→ R given by f Check that f is a linear function. We need to check that f distributes over sums and allows us to pull out scalars. Sums We start with two vectors f while f x1 x2 + f( u1 u2 = ax1 + bx2 + au1 + bu2 . x1 + u1 x2 + u2 x1 u x + u1 and 1 . Their sum is 1 . Now x2 u2 x2 + u2 x1 x2 = ax1 + bx2 . = a(x1 + u1 ) + b(u2 + v2 ) = ax1 + au1 + bx2 + bu2 , So the two computations give the same answer, which says that f distributes over sums. Scaling We start with x1 αx1 and a scalar α; scaling the vector gives . Now we just x2 αx2 compute the two things that are supposed to be equal: f while αf x1 x2 = α(ax1 + bx2 ). αx1 αx2 = aαx1 + bαx2 = α(ax1 + bx2 ), They are equal, so we’ve shown that we can pull out scalars. So we’ve shown that f satisﬁes the two requirements to be a linear function, and we’re done. 2. Still in R2 , look at the special vectors e1 = 1 0 and e2 = . Explain how to get 0 1 any vector from these two, using only adding and scaling operations. Any vector in R2 looks like that 1 0 m =m +n = me1 + ne2 . n 0 1
Never Give an Iguana Viagra – headline, Reuters, January 25, 2007 m for some two numbers m and n. Now just notice n 3. Suppose we know that f : R2 −→ R is a linear function, that f(e1 ) = a and that x f(e2 ) = b. Show that for any vector v = 1 we must have f(v) = ax1 + bx2 . x2 We use problem 2 together with the information that f is linear. f x1 x2 = f(x1 e1 + x2 e2 ) = f(x1 e1 ) + f(x2 e2 ) because we can distribute over sums = x1 f(e1 ) + x2 f(e2 ) because we can pull out scalars = x1 a + x2 b by our deﬁnition of a and b = ax1 + bx2 . So the upshot is that any linear function R2 −→ R must look like the function in problem one. 4. For which choices of a and b is the function f(v) = ax1 + bx2 onto? For which choices is it onetoone? Consider ﬁrst the case a = b = 0. This just gives the constant function equal to zero, so the only possible output is 0, which is “hit” inﬁnitely many times. So in this case f is neither onetoone nor onto. Next, suppose a = 0. Given any real number r, the equation ax1 + bx2 = r translates to x1 = (r − bx2 )/a. So we can choose any value for x2 and there will always be a corresponding solution x1 . Hence this is onto (solutions always exist) but not onetoone (there is one solution for each choice of x2 , hence many solutions). The only remaining case is a = 0, b = 0. But then ax1 + bx2 = r can be rearranged to say x2 = (r − ax1 )/b, and just as before we see that the function is onto but not onetoone. So the upshot is: f is onto for all choices of a and b except for a = b = 0; there are no choices of a and b that make f onetoone. 5. Rodolfo goes to a supermarket where there are only 500 items on offer. His shopping list is an element of R500 , i.e., a vector with 500 entries.
The reality of that which surpasses understanding is not disproved by the recognition that it surpasses understanding. a. Let P be the price function, so that if v is a shopping list then P(v) is the price of buying all the items on your list. Explain why P is a linear function. We can either do this with a formula or by “pure thought.” The entries in Rodolfo’s shopping list are xj , giving the quantity he will buy of the jth product, for j = 1, 2, . . . , 500. If the price of item j is pj , then the price of the list is p1 x1 + p2 x2 + · · · + p500 x500 , which is a linear function. Or we can just say that buying α times as much clearly costs α times as much, etc. b. Price is not Rodolfo’s only concern when he shops. He worries about nutrition too. In particular, there are 12 nutrients he is interested in. So he is interested in the function N from R500 to R12 that computes how much of each of the nutrients you get from that shopping list. Can you describe the function N? Is it a linear transformation? N is indeed a linear transformation: for each of the twelve nutrients the amount you get is a linear function of your shopping list. So N is given by a 12 × 500 matrix. To spell it out: if nij if be the quantity of nutrient i in product j, then the overall quantity of nutrient i is ni1 x1 + ni2 x2 + · · · + ni500 x500 , and that is exactly the effect of multipling the ith row of N by the column [xj ]. 6. Practice matrix multiplication: from the textbook, do problems 1–14 in section 2.3 (page 77). Boring but ultimately just work. 1. An easy matrix multiplication, giving us 46 . 34 2. Another very easy one. We get 4 4 . −8 −8 3. Not deﬁned. ⎤ ⎡ 22 4. ⎣2 0⎦ 74
Innocent people can’t do any good in the world. – Robert Kagan ⎤ ⎡ ab 5. ⎣ c d⎦ 00 6. This is an important one! You get a scalar matrix: (ad − bc) 0 . 0 (ad − bc) ⎡ ⎤ −1 1 0 3 4⎦ 7. ⎣ 5 −6 −2 −4 8. 9. 00 00 00 00 10. 0 1 11. 10 12. This is the only kind of matrix multiplication that involves no adding. The answer is ⎡ ⎤ 123 ⎣2 4 6⎦ . 369 13. h 14. The ones that work are A2 = 22 22 BC = 14 8 2 ⎡⎤ 0 CD = ⎣3⎦ 6 EB = 5 10 15 BD = [6] ⎤ 123 DB = ⎣1 2 3⎦ 123 E2 = [25] ⎡ ⎡ ⎤ −2 −2 −2 1 −2⎦ C2 = ⎣ 4 10 4 −2 ⎡⎤ 5 ⎣5⎦ DE = 5 Few organizations reward incompetence as richly as the United Nations. 7. Long story ommitted. . . I’ll just record the relevant matrices and what they are supposed to mean: A= 10 5 20 10 . 30 20 5 10 Rows: vitamins; columns: pills; entries: units of each vitamin in each pill. ⎤ ⎡ 111 ⎢0 1 2⎥ ⎥ B=⎢ ⎣2 1 0⎦ . 202 Rows: pills; columns: seasons; entries: number of that pill to take that season. a. Compute the product AB. What do the entries in this product represent? AB = 70 35 40 60 55 90 The rows here correspond to vitamins, and the columns to the seasons. The entry in row i column j tells you how much of vitamin i I am getting in season j. So in the winter I am getting 70 units of vitamin D and 60 units of vitamin C, and so on. ⎤ 13 C = ⎣ 2 2⎦ . 23 ⎡ Rows: suppliers; columns: vitamins, entry=cost of vitamin per that supplier. b. What information does the matrix product CA contain? The rows correspond to suppliers, the columns to vitamins. The entry at row i, column j tells me the cost per unit of vitamin j from supplier i. c. What matrix product will tell you how much it costs me to get the pills, according to the supplier and the season? The product CAB has that data: rows correspond to suppliers, columns to seasons, and the entries tell me the daily cost for that season from that supplier. It’s kind of fun to work out what the associative law (CA)B = C(AB) says in the context of my little story about vitamins. My grandfather was born in 1888 and he didn’t have a lifestyle. He didn’t need one: he had a life. – Michael Bywater ...
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This note was uploaded on 10/06/2010 for the course ES ES271 taught by Professor Machaut during the Spring '10 term at Colby.
 Spring '10
 Machaut

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