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Last name: name: 1 Mid term exam 1 (Notes, books, and calculators are not authorized) Show all your work in the blank space you are given on the exam sheet. Always justify your answer. Answers with no justification will not be graded. Question 1: Find the equation of the hyperplane in R3 that passes through P = (2, −1, −5) and is parallel to the hyperplane H 0 determined by the equation 4x1 − 3x2 + 7x3 = 1. The definition of H 0 implies that the vector n = (4, −3, 7) is normal to H 0 . Since H and H 0 are parallel, n = (4, −3, 7) is also a normal vector to H. By definition H is the collection of points X = (x1 , x2 , x3 ) in R3 so that (X − P )·n = 0. This means that 0 =(X − P )·(4, −3, 7) = (x1 − 2, x2 + 1, x3 + 5)·(4, −3, 7) = = 4x1 − 8 − 3x2 − 3 + 7x3 + 35 = 4x1 − 3x2 + 7x3 + 24. The equation of the hyperplane in question is 4x1 − 3x2 + 7x3 = −24. Question 2: Find a parametric representation of the line in R4 that passes through P = (1, 2, 3, 4) in the direction of u = (4, 3, 2, 1). The line in question consists of the points X = (x1 , x2 , x3 , x4 ) in R4 that are such that there is t ∈ R so that (X − P ) = tu. This means that X − P = tu = (4t, 3t, 2t, t). The parametric representation of the line is x1 = 1 + 4t, x2 = 2 + 3t, x3 = 3 + 2t, x4 = 4 + t. or L = {(1 + 4t, 2 + 3t, 3 + 2t, 4 + t) ∈ R4 , t ∈ R}. 2 Mid term exam 1, October 16, 2014 Question 3: Prove the complex Cauchy-Schwarz inequality: i.e., |Re(hu, vi)| ≤ kuk`2 kvk`2 , for all u, v ∈ Cn and all positive integer n. (Hint: consider the polynomial Re(htu + v, tu + vi) with t ∈ R, and use the observation that Re(u · v̄) = Re(v · ū)). We follow the hint and consider the polynomial Re(htu + v, tu + vi) with t ∈ R. Then 0 ≤ ktu + vk`2 Re(htu + v, tu + vi) = Re(t2 u · ū + tu · v̄ + tv · ū + v · v̄) = Re(t2 kuk`2 + tu · v̄ + tv · ū + kvk`2 ) = t2 kuk`2 + tRe(u · v̄ + v · ū) + kvk`2 = t2 kuk`2 + 2tRe(u · v̄) + kvk`2 . The polynomial t2 kuk2`2 + 2tRe(u · v̄) + kvk2`2 is non-negative for all values of t. This means that the discriminant is non-positive: 4Re(u · v̄) − 4kuk`2 kvk`2 ≤ 0. In conclusion ∀u, v ∈ Cn . Re(u · v̄) ≤ kuk`2 kvk`2 , The same inequality obviously holds by replacing v by −v: −Re(u · v̄)| ≤ kuk`2 kvk`2 , ∀u, v ∈ Cn . |Re(u · v̄)| ≤ kuk`2 kvk`2 , ∀u, v ∈ Cn . In conclusion Question 4: Find a 2×2 real-valued matrix P whose columns are unit orthogonal vectors and whose second column is a multiple of (1, −2)T . The second column of P is λ(1, −2) where λ ∈ R. The norm of this vector is λ2 (1 + 4) = 5λ2 . Since this is a unit vector, we infer that λ = ± √15 . Let us choose λ = √15 . The first column of P is (a, b)T , i.e., # " √1 a 5 P = b − √25 with a2 + b2 = 1 and a √15 − b √25 = 0. This means a = 2b and b2 (1 + 4) = 1, i.e., a = ± √25 and b = ± √15 . In conclusion we can choose 1 2 P =√ 5 1 1 . −2 Last name: name: 1 0 Question 5: Let E = 0 1 0 −2 R2 and R3 are row vectors. (a) 3 0 R1 0 and let A = R2 be an arbitrary 3 × 3 matrix, where R1 , 1 R3 Compute EA E is the matrix on the elementary row operation that consists of replacing the second row of a matrix by itself minus three times the first row: R1 R2 EA = −2R2 + R3 (b) Give the inverse of E E because 1 0 0 0 1 2 0 1 0 EA = 0 1 0 −1 1 = 0 0 0 1 2 0 0 1 0 R1 R1 = = A. 0 R2 R2 1 −R2 + R3 2R2 − 2R2 + R3 0 1 2 Using A = I in the above equality means that 1 0 0 1 0 2 0 0 E = I 1 Question 6: Let u be a unit vector in Cn , i.e., kuk2`2 = ūT u = 1. (a) Prove that H = 1 − 2uūT is Hermitian. (Recall that H is Hermitian iff H = H̄ T .) The definition of H implies that T H̄ T = (I − 2uūT ) = (I − 2ūuT )T = (I − 2uūuT ) = H. (b) Prove that H = I − 2uūT is unitary. (Recall that H is unitary iff H H̄ T = I.) Using the fact that H is Hermitian, we infer that H H̄ T = HH = (I − 2uūT )(I − 2uūT ) = I − 4uūT + 4uūT uūT . Now using that u is a unit vector, ūT u = 1, we infer that H H̄ T = I − 4uūT + 4uūT = I, i.e., H is Hermitian. 4 Mid term exam 1, October 16, 2014 Question 7: Are the following functions f1 (t) = 1 − 3t + 2 sin(t), f2 (t) = 2 − 4t − sin(t), f3 (t) = 1 − 5t + 7 sin(t) linearly independent in V := span{1, t, sin(t)}? Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 f1 + x2 f2 + x3 f3 = 0. Then x1 f1 (t) + x2 f2 (t) + x3 f3 (t) = x1 + 2x2 + x3 + (−3x1 − 4x2 − 5x3 )t + (2x1 − x2 + 7x3 ) sin(t) = 0. This means that X solves 1 −3 2 2 −4 −1 1 −5 X = 0. 7 Let us reduce the matrix of the linear system in echelon form 1 2 1 1 2 1 1 −3 −4 −5 ∼ 0 2 −2 ∼= 0 2 −1 7 0 −5 5 0 2 1 2 −2 . 0 0 There are only two pivots. There is one free variable. This means that the solution set of the above linear system is not {0}. There is some nonzero vector X so that x1 f1 + x2 f2 + x3 f3 = 0. This means that the functions f1 , f2 , f3 are linearly dependent. 4 −3 2 −1 16 −9 6 −3 Question 8: Compute the LU factorization of A = 12 3 0 0 . Give all the details. 8 3 6 −2 4 −3 2 −1 1 0 0 0 4 1 0 0 0 3 −2 1 A= 3 4 1 0 0 0 2 1 0 0 0 1 2 3 4 1 Last name: name: 5 Question 9: Express the function v(t) = e2t + 4et − 3 as a linear combination of z1 (t) = e2t − 2et + 5, z2 (t) = 2e2t − 3et , z3 (t) = et + 3. We look for (x1 , x2 , x3 ) ∈ R3 so that v = x1 z1 + x2 z2 + x3 z3 . This means that the following holds for all t ∈ R: e2t + 4et − 3 = x1 (e2t − 2et + 5) + x2 (2e2t − 3et ) + x3 (et + 3) = e2t (x1 + 2x2 ) + et (−2x1 − 3x2 + x3 ) + 1(5x1 + 3x3 ). This implies x1 + 2x2 = 1, We can write this form: 1 2 −2 −3 5 0 −2x1 − 3x2 + x3 = 4 5x1 + 3x3 = −3. system in the form of an augmented matrix and compute the reduced echelon 0 1 1 1 4 ∼ 0 3 −3 0 2 1 −10 0 1 1 1 6 ∼ 0 3 −8 0 2 1 0 0 1 13 1 1 6 ∼ 0 52 0 0 1 0 0 0 1 −3 2 4 This means that v = −3z1 + 2z2 + 4z3 . Question 10: (a) Let A be a n × n real-valued matrix. Assume that X T AX ≥ −5kXk2`2 for all X ∈ Rn . Prove that 7I + A is invertible. (b) Let A be a n × n complex-valued matrix. Assume that Re(X̄ T AX) ≥ −5kXk2`2 for all X ∈ Cn . Prove that 7I + A is invertible. 6 Mid term exam 1, October 16, 2014 Question 11: Consider the two-variate polynomials Q1 (x, y) = (1 − x)(1 − y), Q2 (x, y) = x(1 − y), Q3 (x, y) = y(1 − x), Q4 (x, y) = xy. (a) Show that the set S = {Q1 , Q2 , Q3 , Q4 } is linearly independent. Let x1 , x2 , x3 , x3 such that x1 Q1 + . . . + x4 Q4 = 0. Then 0 = x1 (1 − x − y + xy) + x2 (x − xy) + x3 (y − xy) + x4 xy = x1 + (−x1 + x2 )x + (−x1 + x3 )y + (x1 − x2 − x3 + x4 ). This means that x1 = 0, −x1 + x2 = 0, −x1 + x3 = 0, x1 − x2 − x3 + x4 = 0. Then clearly x1 = x2 = x3 = x4 = 0, i.e., the set S is linearly independent. (b) Is the polynomial Q(x, y) = 1 in span(S)? Let us try to find x1 , x2 , x3 , x3 such that 1 = x1 Q1 + . . . + x4 Q4 This implies that 1 = x1 (1 − x − y + xy) + x2 (x − xy) + x3 (y − xy) + x4 xy = x1 + (−x1 + x2 )x + (−x1 + x3 )y + (x1 − x2 − x3 + x4 ). Then x1 = 1, −x1 + x2 = 0, −x1 + x3 = 0, x1 − x2 − x3 + x4 = 0. Then clearly x1 = x2 = x3 = x4 = 1. The means that 1 = Q1 (x, y)+Q2 (x, y)+Q3 (x, y)+Q4 (x, y), i.e., 1 ∈ span(S)