Solutions to Assignment 3

Transcription

1 Solutions to Assignment Math 17, Fall Let T : R R be a linear transformation with standard matrix A = [a 1 a ], [ where ] a 1 and a are shown in the figure. Using the figure, draw the image of under the transformation T. x a 1 x 1 a ([ ]) [ ] [ ] The key to this problem is realizing that T = A = [a 1 a ] = a 1 + a. This is what it means for A to be the standard matrix of T. So we need to draw the vector a 1 + a. That is simply: 1

2 x a 1 a 1 T a 1 a 1 a x 1 a 1.9. Let T : R R be a linear transformation such that T Find x such that T (x) = 4. 9 ([ x1 x ]) x 1 x = x 1 + x. x 1 x This problem isn t hard if we write down the standard matrix of T. So, ([ ]) x 1 x 1 1 [ ] x1 T = x x 1 + x = x 1 + x = x1, x x 1 x and it is clear that the standard matrix of T is 1 A =. So now we put the augmented matrix 1 A = 4 9

3 Let is row reduced echelon form. After the dust clears, we get the matrix The solution to this linear system is x = T (x) = 4. 9 [ ] 5, that is, A = be the standard matrix of T : R 5 R 5. Is T onto? [ ] 5 is an x such that According to Theorem 1 page 89, this map is onto if and only if the columns of A span R 5. In turn, by Theorem 4, page 4, we know that A spans R 5 if and only if A has 5 pivot positions. So we put the monster in reduced row echelon form (row echelon form is enough, but my calculator gives the answer reduced). We find the matrix As this fails to have 5 pivot positions, we conclude that T is not onto One serving (8 g) of Kellogg s Cracklin Oat Bran supplies 110 calories, g of protein, 1 g of carbohydrate, and g of fat. One serving of Kellogg s Crispix supplies 110 calories, g of protein, 5 g of carbohydrate, and.4 g of fat. (a) Set up a matrix B and a vector u such that Bu gives the amounts of calories, protein, carbohydrate, and fat contained in a mixture of three servings of Cracklin Oat Bran and two servings of Crispix. Let u 1 be the number of servings of Cracklin Oat Bran and u be the number of servings of Crispix in our mixture. Then there are 110u u calories in any mixture, u 1 + u g of protein, 1u 1 + 5u g of carbohydrate, and u 1 +.4u g of fat. This corresponds to u 1 +

4 u, that is, So the matrix is [ u1 u ] B = For servings of Cracklin Oat [ Bran ] and servings of Crispix, we would multiply B by the vector u =. (b) Suppose that you want a cereal with more protein than Crispix but less fat than Cracklin Oat Bran. It is possible for a mixture of the two cereals to supply 110 calories,.5 g of protein, 4 g of carbohydrate, and 1 g of fat? If so, what is the mixture? 110 Solve the system Bu = augmented matrix is The row reduced echelon form of the We see that the system is inconsistent, so there is no mixture that works..1.4 Compute A 5I and (5I )A, when 9 A = Well, A 5I = =

5 Also, (5I )A = = Suppose the first two columns, b 1 and b, of B are equal. What can you say about the columns of AB (if AB is defined)? Why? Well, we can write B = [b 1 b 1 b b n ], and then AB = [Ab 1 Ab 1 Ab Ab n ]. One can see that the first two columns of AB are equal..1.4 Suppose AD = I m (the m m identity matrix). Show for any b in R m, the equation Ax = b has a solution. Explain why A cannot have more columns then rows. Let b be any vector in R m. Then Db is a vector as well, and A(Db) = (AD)b = I m b = b, so Ax = b always has a solution. By Theorem 4, page 4, this means that A has a pivot position in every row. If there were more rows then columns, then not each row could have a pivot (because there is at most one pivot per column). So there must be at least as many columns as rows...14 Suppose AB = AC, where B and C are n p matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible? We may multiply both sides of the equality above by A. The result is A AB = A AC. Simplifying both sides we [ obtain ] B = [ C. ] This is not true in general, however. Suppose that A =, B =, and C = [ ] Note that A is not invertible (see Theorem 4 of this chapter). Then 0 1 [ ] 0 0 AB = = AC, but B C Suppose A and B are n n, B is invertible, and AB is invertible. Show that A is invertible. We know that B and (AB) exist. Let C = AB. Then CB = A, after a little painless algebra. But we have noted already (see page 1) that the product of invertible things is invertible. Both C and B are invertible (see Theorem 6 for the second fact), and thus A is invertible. In this case, A = BC... Explain why the columns of an n n matrix A span R n when A is invertible. If A is invertible, then Ax = b has a solution for all b R n (Theorem 5). But by Theorem 4, page 4, this occurs if and only if the columns of A span R n...16 Is it possible for a 5 5 matrix to be invertible when it columns do not span R 5? No. If the matrix in question is invertible, then because it is square, it columns must span R 5 by the Invertible Matrix Theorem (look at (a) and (h) of that theorem). 5

6 ..18 If C is 6 6 and the equation Cx = v is consistent for every v in R 6, is it possible that for some v, the equation Cx = v has more than one solution? Why or why not? By the Invertible Matrix Theorem, Cx = v is consistent for every v in R 6 if and only if C is invertible. Theorem 5, page 10, tells us that if this is so, then the solution to Cx = v is unique (it is C v)...6 Let T be a linear transformation that maps R n onto R n. Show that T exists and maps R n onto R n. Is T also one-to-one? Let A be the standard matrix of T. So x Ax is onto, and by the Invertible Matrix Theorem A is invertible. Then by Theorem 9, T exists with standard matrix A. But by Theorem 6 page 11, A is invertible, so that again by the Invertible Matrix Theorem we know that x A x is onto (and one-to-one). This means that T is onto (and one-to-one). 6