Linear Algebra Exercises-n-Answers.pdf

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24 Linear Algebra, by Hefferon and then as required. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⃗u ( k⃗v + m ⃗w ) u 1 ⎜ = ⎝ ⎟ . ⎠ ( kv 1 mw 1 ⎜ ⎝ ⎟ ⎜ . ⎠ + ⎝ ⎟ . ⎠ ) u n kv n mw n ⎛ ⎞ ⎛ ⎞ u 1 kv 1 + mw 1 ⎜ ⎟ ⎜ ⎟ = ⎝ . ⎠ ⎝ . ⎠ u n kv n + mw n = u 1 (kv 1 + mw 1 ) + · · · + u n (kv n + mw n ) = ku 1 v 1 + mu 1 w 1 + · · · + ku n v n + mu n w n = (ku 1 v 1 + · · · + ku n v n ) + (mu 1 w 1 + · · · + mu n w n ) = k(⃗u ⃗v) + m(⃗u ⃗w) One.II.2.38 For x, y ∈ R + , set ⃗u = (√ ) x √ y ⃗v = (√ ) √ y x so that the Cauchy-Schwartz inequality asserts that (after squaring) as desired. ( √ x √ y + √ y √ x) 2 ≤ ( √ x √ x + √ y √ y)( √ y √ y + √ x √ x) (2 √ x √ y) 2 ≤ (x + y) 2 √ xy ≤ x + y 2 One.II.2.39 This is how the answer was given in the cited source. The actual velocity ⃗v of the wind is the sum of the ship’s velocity and the apparent velocity of the wind. Without loss of generality we may assume ⃗a and ⃗ b to be unit vectors, and may write ⃗v = ⃗v 1 + s⃗a = ⃗v 2 + t ⃗ b where s and t are undetermined scalars. Take the dot product first by ⃗a and then by ⃗ b to obtain s − t⃗a ⃗ b = ⃗a (⃗v 2 − ⃗v 1 ) s⃗a ⃗ b − t = ⃗ b (⃗v 2 − ⃗v 1 ) Multiply the second by ⃗a ⃗ b, subtract the result from the first, and find s = [⃗a − (⃗a ⃗ b) ⃗ b] (⃗v 2 − ⃗v 1 ) 1 − (⃗a ⃗ b) 2 . Substituting in the original displayed equation, we get One.II.2.40 We use induction on n. In the n = 1 base case the identity reduces to and clearly holds. ⃗v = ⃗v 1 + [⃗a − (⃗a ⃗ b) ⃗ b] (⃗v 2 − ⃗v 1 )⃗a 1 − (⃗a ⃗ b) 2 . (a 1 b 1 ) 2 = (a 1 2 )(b 1 2 ) − 0 For the inductive step assume that the formula holds for the 0, . . . , n cases. We will show that it
Answers to Exercises 25 then holds in the n + 1 case. Start with the right-hand side ( ∑ 2 a )( ∑ 2 j b ) ∑ ( ) 2 j − ak b j − a j b k 1≤j≤n+1 1≤j≤n+1 1≤j≤n 1≤k
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