Linear Algebra (Math 311) – Final

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From the last matrix in the computation we see that the matrix is nonsingular, and we conclude that the first, second, and fourth row are linearly independent. 3. The first three columns of A are linearly independent because that is where B has the leading 1’s. 4. The orthogonal complement to the row<strong>–</strong>space is the null space of the matrix. The last two ccordinates are free. We obtain a basis by setting either x4 = 1 and x5 = 0 or x4 = 0 and x5 = 1. With these choices we get the basis ⎧⎛ ⎞ 6 ⎪⎨ ⎜ ⎜−4 ⎟ ⎜ 1 ⎟ ⎝ ⎪⎩ 1 ⎠ 0 , ⎛ ⎞⎫ −75 ⎜ 47 ⎟⎪⎬ ⎟ ⎜ −9 ⎟ ⎝ 0 ⎠ ⎪⎭ 4 5. We solved part 5. as part 2. of the problem. Delete the last two columns of the matrix, as well as row three and five. The result is a nonsingular matrix. There is not larger nonsingular minor, because is size will be the rank of the matrix, and that is 3. 6. As stated in 4. the nullspace of the matrix is the orthogonal complement of its rowspace. The nullity is its dimension, and it is the number of columns minus the rank of the matrix. The nullity is 5 − 3 = 2. Problem 2. [10 points] Consider the vector space P2 of polynomials of degree at most 2 with the basis consisting of the vectors v1 = 1, v2 = 1 + t, and v3 = 1 + t + t 2 . Differentiation defines a linear map D : P2 → P2. Find the matrix of D with respect to the given basis. Solution: We apply D to the basis elements, making sure that the results are given in terms of the basis again. We find: D(v1) = 0, D(v2) = 1 = v1, and D(v2) = 1 + 2t = −1 + 2(1 + t) = −v1 + 2v2. Accordingly, the matrix for D is ⎛ ⎞ 0 1 −1 ⎝0 0 2 ⎠ 0 0 0 Problem 3. [20 points] Let P2 be as in Problem 2. Set S = {1 − t 2 } and T = {1 + t, 1 − t 2 , t + t 2 , 1 + 2t, 1 + t + t 2 } 2
1. Is it possible to add elements to S until you have a basis of P2? Explain why! If possible, do it. 2. Is it possible to omit elements of T until you have a basis of P2? Explain why! If possible, do it. Solution: Observe that P2 is of dimension 3, so we are looking for bases consisting of three elements. 1. Because S is linearly independent, we can add elements, keeping the set linearly independent, until the set spans and we have a basis. Here is a set that works, obviously: {1 − t 2 , 1, t} 2. Note that T spans P2 and contains a basis because 1 = (1 + t + t 2 ) − (t + t 2 ) t = (1 + 2t) − (1 + t) t 2 = (1 + t + t 2 ) − (1 + t) To cut down T to a basis, we use the first element, it is nonzero. We keep the second element, because the first two elements are linearly independent, none of the elements is a multiple of the other one. Note that the third element is the difference of the first two, so discard it. It is easy to see that the fourth element is independent of the first two, so we keep it. By now we have three linearly independent elements selected from T , and they form a basis: {1 + t, 1 − t 2 , 1 + 2t, } Problem 4. [15 points] Find the determinants of the matrices ⎛ ⎞ ⎛ ⎞ 1 2 3 1 3 0 1 ⎜ A = ⎝1 0 4⎠ and B = ⎜0 1 2 5 ⎟ ⎝1 2 1 1⎠ 2 1 5 0 7 1 2 Solution: Using the standard formula for 3 × 3 matrices, we find det A = 1 − 12 = −11. We may apply elementary row operations (recalling how they affect the determinate) to bring the matrix into triangular form, and then get the determinante as the product of the diagonal elements. Here is one way, without 3
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Page 5 and 6: Solution: A norm on a vector space

Linear Algebra (Math 311) – Final

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