Linear Algebra Exercises-n-Answers.pdf

130 Linear Algebra, by Hefferon
then the vector ⃗w whose components are v 2 , −v 1 , 0, . . . , 0 is perpendicular to ⃗v. In either case, observe
that ⃗v + ⃗w does not equal ⃗v, and that ⃗v is the projection of ⃗v + ⃗w into l.
(⃗v + ⃗w) ⃗v
· ⃗v = ( ⃗v ⃗v ⃗w ⃗v ) ⃗v ⃗v
+ · ⃗v = · ⃗v = ⃗v
⃗v ⃗v ⃗v ⃗v ⃗v ⃗v ⃗v ⃗v
We can dispose of the remaining n = 0 and n = 1 cases. The dimension n = 0 case is the trivial
vector space, here there is only one vector and so it cannot be expressed as the projection of a different
vector. In the dimension n = 1 case there is only one (nondegenerate) line, and every vector is in it,
hence every vector is the projection only of itself.
Three.VI.1.15 The proof is simply a calculation.
⃗v ⃗s ⃗s |⃗v ⃗s | |⃗v ⃗s |
‖ · ⃗s ‖ = |⃗v | · ‖⃗s ‖ = · ‖⃗s ‖ =
⃗s ⃗s ⃗s ⃗s ‖⃗s ‖
2
‖⃗s ‖
Three.VI.1.16 Because the projection of ⃗v into the line spanned by ⃗s is
⃗v ⃗s
⃗s ⃗s · ⃗s
the distance squared from the point to the line is this (a vector dotted with itself ⃗w ⃗w is written ⃗w 2 ).
⃗v ⃗s
‖⃗v −
⃗s ⃗s · ⃗s ⃗v ⃗s ⃗s
⃗s
‖2 = ⃗v ⃗v − ⃗v ( · ⃗s) − (⃗v · ⃗s ) ⃗v + (⃗v · ⃗s )2
⃗s ⃗s ⃗s ⃗s ⃗s ⃗s
⃗v ⃗s
⃗s
= ⃗v ⃗v − 2 · ( ) · ⃗v ⃗s + (⃗v ) · ⃗s ⃗s
⃗s ⃗s ⃗s ⃗s
= (⃗v ⃗v ) · (⃗s ⃗s ) − 2 · (⃗v ⃗s )2 + (⃗v ⃗s ) 2
⃗s ⃗s
(⃗v ⃗v )(⃗s ⃗s ) − (⃗v ⃗s )2
=
⃗s ⃗s
Three.VI.1.17 Because square root is a strictly increasing function, we can minimize d(c) = (cs 1 −
v 1 ) 2 +(cs 2 −v 2 ) 2 instead of the square root of d. The derivative is dd/dc = 2(cs 1 −v 1 )·s 1 +2(cs 2 −v 2 )·s 2 .
Setting it equal to zero 2(cs 1 − v 1 ) · s 1 + 2(cs 2 − v 2 ) · s 2 = c · (2s 2 1 + 2s 2 2) − (v 1 s 1 + v 2 s 2 ) = 0 gives the
only critical point.
c = v 1s 1 + v 2 s 2 ⃗v ⃗s
=
s 12 + s
2 2 ⃗s ⃗s
Now the second derivative with respect to c
d 2 d
dc 2 = 2s 1 2 2
+ 2s 2
is strictly positive (as long as neither s 1 nor s 2 is zero, in which case the question is trivial) and so the
critical point is a minimum.
The generalization to R n is straightforward. Consider d n (c) = (cs 1 − v 1 ) 2 + · · · + (cs n − v n ) 2 , take
the derivative, etc.
Three.VI.1.18 The Cauchy-Schwartz inequality |⃗v ⃗s | ≤ ‖⃗v ‖ · ‖⃗s ‖ gives that this fraction
⃗v ⃗s ⃗s |⃗v ⃗s | |⃗v ⃗s |
‖ · ⃗s ‖ = |⃗v | · ‖⃗s ‖ = · ‖⃗s ‖ =
⃗s ⃗s ⃗s ⃗s ‖⃗s ‖
2
‖⃗s ‖
when divided by ‖⃗v ‖ is less than or equal to one. That is, ‖⃗v ‖ is larger than or equal to the fraction.
Three.VI.1.19 Write c⃗s for ⃗q, and calculate: (⃗v c⃗s/c⃗s c⃗s ) · c⃗s = (⃗v ⃗s/⃗s ⃗s ) · ⃗s.
Three.VI.1.20
(a) Fixing
(
1
⃗s =
1)
as the vector whose span is the line, the formula gives this action,
( ) (
x 1
( (
x y 1)
1
↦→ ( ) ( · =
y)
1 1 1)
1 1) x + y ( ( )
1 (x + y)/2
· =
2 1)
(x + y)/2
which is the effect of this matrix. ( 1/2
) 1/2
1/2 1/2
(b) Rotating the entire plane π/4 radians clockwise brings the y = x line to lie on the x-axis. Now
projecting and then rotating back has the desired effect.