Carleton College, winter 2013 Math 232, Solutions to review problems and practice midterm 2 Prof. Jones 15. T 17. F 38. T 21. F 26. T 22. T 27.

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1 Carleton College, winter 23 Math 232, Solutions to review problems and practice midterm 2 Prof. Jones Solutions to review problems: Chapter 3: 6. F 8. F. T 5. T 23. F 7. T 9. F 4. T 7. F 38. T Chapter 4:. F 6. F 3. F 2. F 26. T 32. T 44. F 55. T 2. T 7. T 6. T 22. T 27. T 34. F 45. T 64. F 4. T. T 7. F 24. T 28. T 37. T 5. T 5. T 2. F 9. F 25. F 3. F 43. T 53. T Solutions to practice exam:. Find a bases for the kernel and image of the following matrix: 2 A = To find a basis for the kernel, we need to row reduce the augmented matrix 2 3, 4 2 and we get the reduced row-echelon form /3 /6 /2, Thus the free variables are x 2, x 4, and x 5, and we set x 2 = t, x 4 = t 2, and x 5 = t 3. The set of solutions is then x x 2 t 3 2 x 3 x 4 = t 6 t 2 t 2 3 t 2 x 5 t 3

2 Separating the arbitrary constants t, t 2, t 3 into their own vectors then yields a basis of /3, /6, /2. As for the image, the above reduced row-echelon form shows that the non-redundant columns are the first and the third, so a basis for the image is 2 3, Find a basis for the following subspace of P 4. What is the dimension of W? Note that W is the same as W = {p(x P 4 p( = p( = }. {a 4 x 4 + a 3 x 3 + a 2 x 2 + a x + a P 4 a 4 + a 3 + a 2 + a + a =, a 4 a 3 + a 2 a + a = }. There are different ways to do this problem, but the most systematic is to find all solutions to the system of equations a +a +a 2 +a 3 +a 4 = a a +a 2 a 3 +a 4 =. So we put the augmented matrix [ ] in reduced row-echelon form. We get [ Thus our free variables are a 2, a 3, and a 4. Set a 2 = t, a 3 = t 2, and a 4 = t 3. The set of solutions is then a t t 3 a a 2 a 3 = t 2 t t 2 a 4 t 3 2 ]

3 Separating the arbitrary constants t, t 2, t 3 into their own vectors then yields a basis of,,. But we need to re-convert these to polynomials, since W is a subspace of P 4, and hence any basis for it should consist of elements of P 4. So a basis for W is and W has dimension 3. { + x 2, x + x 3, + x 4 }, 3. Determine whether the following mappings are linear transformations. Either prove that a given map is linear or give a counterexample to show it s not linear. (a T : R 2 R 3 defined by T ((x, x 2 = (2x, x + 4, 5x 2 Not linear since 2T ((, = 2(, 4, = (, 8, but T (2(, = T ((, = (, 4,. (b T : P 2 P 3 defined by T (a 2 x 2 + a x + a = a x 3 + (a a x 2 + 3a 2 (/2a Linear. Let u = u 2 x 2 + u x + u and v = v 2 x 2 + v x + v. Then T ( u + v = (u + v x 3 + (u + v u v x 2 + 3(u 2 + v 2 (/2(u + v = (u x 3 + (u u x 2 + 3u 2 (/2u + (v x 3 + (v v x 2 + 3v 2 (/2v = T ( u + T ( v A similar calculation shows that T (c u = ct ( u. 4. Let V be a finite-dimensional vector space, and let S be a linearly independent subset of V. Let S be a proper subset of S (this means that S S and S S. Prove that S cannot be a basis for V. [Hint: use Theorem ] Suppose that the dimension of V is n. Since S is a linearly independent subset of V, by Theorem 3.3.4(a, S can contain at most n vectors. Since S is a proper subset of S, it must contain strictly fewer than n vectors. But by Theorem 3.3.4(b, at least n vectors are required to span V, and hence S cannot span V. Therefore S cannot be a basis for V. 5. (a Consider the mapping T : R 2 2 R 2 2 defined by ( ( a b a + b T = c d b + 2d 3c c b d + 4a. 3

4 Prove that T is a linear transformation. Let Then T ( u + v = = ( u u u = 2 u 2 u 22 ( v v v = 2 v 2 v 22 ( u + v + u 2 + v 2 u 2 + v 2 (u 2 + v 2 u 2 + v 2 + 2(u 22 + v 22 3(u 2 + v 2 u 22 + v (u + v ( ( u + u 2 u 2 u 2 v + + v 2 v 2 v 2 u 2 + 2u 22 3u 2 u u v 2 + 2v 22 3v 2 v v = T ( u + T ( v A similar calculation shows that T (c u = ct ( u. (b Given the basis α = {( (, (, give the matrix [T ] α of T with respect to the basis α. (, } of R 2 2, T ( = ( 4 Thus the first column of [T ] α is ( = ( + 4 Proceeding similarly with the other columns, we get [T ] α = ( + ( The mapping T : R 2 P 2 given by T ((a, a 2 = (a + a 2 x 2 + a 2 x + a is a linear transformation. (a Prove that α = {(, 2, (, } is a basis for R 2 and β = {x 2 + 2, x 2 + x, } is a basis for P 2. First note that α is a basis for R 2 since α is linearly independent (it is a set of two vectors and neither is a multiple of the other and the dimension of R 2 is two. Similarly, one can check 4

5 that β is linearly independent by assuming that c (x c 2 (x 2 + x + c 3 ( =, setting up a system of equations involving c, c 2, and c 3, and showing that c = c 2 = c 3 =. Since P 2 (R has dimension 3 this shows that β is a basis, by Theorem 3.3.4(c. (b Find the matrix [T ] β α To compute the first column of [T ] β α, we find T ((, 2 and write it in β-coordinates. We have and thus the first column of [T ] β α is T ((, 2 = 3x 2 + 2x + = (x (x 2 + x ( 2 Note that to find the β-coordinates of 3x 2 +2x+, you can either eyeball a solution as follows: only x 2 + x involves x, and so we must have 2(x 2 + x, and then the only other term with x 2 is x 2 + 2, so we must add on (x to get 3x 2, then we have a constant term of 2, so we must subtract (. Or you can take the more systematic approach of writing which leads to the system of equations which you can solve. 3x 2 + 2x + = a (x a 2 (x 2 + x + a 3 (, a 2 +a = 3 a 2 = 2 a 3 +2a = In the same way we can find the second column of [T ] β α, which gives [T ] β α = 2 (c What is the dimension of Ker(T? Find a basis for Ker(T. We put the augmented matrix In reduced row-echelon form, which gives 2 5

6 There are no free variables, and hence ker T = { }. So ker T is zero-dimensional, and has basis { }. (d Since the dimension of ker T is zero, by the rank-nullity theorem we know that the dimension of the image of T must equal the dimension of R 2, which is two. A basis is 2,. 6