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quiz2_1806_sol_s05

# quiz2_1806_sol_s05 - b = QQ T b(c The error vector e = b...

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Exam 2, Friday April 1st, 2005 Solutions Question 1. The vector a 1 can be any non-zero positive multiple of q 1 . The vector a 2 can be any multiple of q 1 plus any non-zero positive multiple of q 2 : a 1 = cq 1 , with c, c 1 > 0 . a 2 = c 1 q 1 + c 2 q 2 Question 2. We want to Fnd the least squares solution to the equation ax = b and we know that it is enough to multiply both sides by a T and solve the resulting system: a · b T T a ax = a b = x = a · a Question 3. The vectors ( 1 , 1 , 0) T and ( 1 , 0 , 1) T form a basis for the subspace x + y + z = 0. Let A be the matrix whose columns are the two vectors found above. Thus the projection matrix P onto the subspace x + y + z = 0 is 1 1 ± 1 ± 1 1 0 P = A A T A 1 A T = 1 0 ² 2 1 = 1 2 1 0 1 0 1 1 1 ± ⎝ ± 1 2 1 1 1 0 = 1 0 ² = 3 1 2 1 0 1 0 1 1 1 ± 1 1 2 1 = 1 0 ² = 3 1 1 2 0 1 2 1 1 1 = 1 2 1 ² 3 1 1 2 The projection of (1 , 2 , 6) T onto the plane x + y + z = 0 is thus simply 1 2 p = P 2 ² = 1 ² 6 3 Question 4. Looking at the Frst row of A we deduce that 1 1 0 det A = det 1 2 3 ² = 9 6 = 3 1 3 9 1

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Of course, det( A 1 ) = 1 . Finally 3 0 1 0 A 1 C 21 1 ( 9) = = det 2 1 3 ± = = 3 12 det A 3 3 3 1 9 Question 5. (a) The column space of QQ T is at most two dimensional, since the matrix QQ T is 4 × 4, it cannot have rank four. Thus det QQ T = 0. Similarly, the matrix [ Q Q ] has dependent columns, and therefore det[ Q Q ] = 0. (b) Using the projection formula, p = Q ( Q T Q ) 1 Q T b = QIQ T
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Unformatted text preview: b = QQ T b (c) The error vector e = b − p is contained in the left null-space of Q , the nullspace of Q T . To check this, we compute ⎛ ⎜ Q T e = Q T b − QQ T b = Q T b − Q T QQ T b = Q T b − IQ T b = Question 6. The product P 2 P 1 is projection onto the column space of P 1 , followed by the projection onto the column space of P 2 . Since the column space of P 2 contains the column space of P 1 , the second projection does not change the vectors anymore. Thus ⎞ ⎞ ⎞ − 1 ⎞ 1 1 1 2 1 ⎟ 2 ² ⎟⎛ ⎜ ⎟ 2 ²² ⎛ 2 ² ⎜ 1 ⎟ 2 4 ⎟ ² ⎟ 1 ⎟ ² P 2 P 1 = P 1 = 2 1 ⎟ ²² 1 2 1 = ⎠ ± ⎠ ⎠ ±± 6 ⎠ ± 1 1 1 2 1 2...
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quiz2_1806_sol_s05 - b = QQ T b(c The error vector e = b...

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