Linear Algebra Exercises-n-Answers.pdf

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