5.pdf

(434 KB) Pobierz
673462036 UNPDF
Chapter 5
ISM: Linear Algebra
Chapter 5
5.1
1.kvk=
p
7 2 + 11 2 =
p
49 + 121 =
p
17013:04
2.kvk=
p
2 2 + 3 2 + 4 2 =
p
4 + 9 + 16 =
p
295:39
3.kvk=
p
2 2 + 3 2 + 4 2 + 5 2 =
p
4 + 9 + 16 + 25 =
p
547:35
4. = arccos
uv
kukkvk = arccos
7+11
p
2
p
170 = arccos
18
p
340
0:219 (radians)
5. = arccos
kukkvk = arccos 2+6+12
p
14
p
0:122 (radians)
29
6. = arccos
kukkvk = arccos 23+810
p
10
p
1:700 (radians)
54
7. Use the fact that uv =kukkvkcos , so that the angle is acute if uv > 0, and obtuse if
uv < 0. Since uv = 1012 =2, the angle is obtuse.
8. Since uv = 424 + 20 = 0, the two vectors enclose a right angle.
9. Since uv = 34 + 53 = 1, the angle is acute (see Exercise 7).
10. uv = 2 + 3k + 4 = 6 + 3k. The two vectors enclose a right angle if uv = 6 + 3k = 0,
that is, if k =2.
11. a. n = arccos
kukkvk = arccos
uv
1 p
n
2 = arccos
1 p
2 = 4 (= 45 )
3 = arccos
1 p
3
0:955 (radians)
4 = arccos 2 = 3 (= 60 )
b. Since y = arccos(x) is a continuous function,
n!1 n = arccos
n!1 1 p
n
= arccos(0) = 2 (= 90 )
12.kv + wk 2 = (v + w)(v + w) (by hint)
=kvk 2 +kwk 2 + 2(vw) (by denition of length)
kvk 2 +kwk 2 + 2kvkkwk(by Cauchy-Schwarz)
234
uv
uv
lim
lim
ISM: Linear Algebra
Section 5.1
= (kvk+kwk) 2 , so that
kv + wk 2 (kvk+kwk) 2
Taking square roots of both sides, we nd thatkv + wkkvk+kwk, as claimed.
13. Figure 5.1 shows thatkF 2 + F 3 k= 2 cos
2
kF 2 k= 20 cos
2
.
It is required thatkF 2 + F 3 k= 16, so that 20 cos
2
= 16, or = 2 arccos(0:8)74
.
Figure 5.1: for Problem 5.1.13.
14. The horizontal components of
F 1 and
F 2 arekF 1 ksin andkF 2 ksin , respectively (the
F 3 is zero).
horizontal component of
Since the system is at rest, the horizontal components must add up to 0, so thatkF 1
ksin +
kF 2
ksin = 0 orkF 1
ksin =kF 2
ksin or
kF 1
k = sin
sin .
kF 2
To nd
EB , note that EA = ED tan and EB = ED tan so that
EA
EB
=
tan
tan
=
sin
sin
cos
cos
=
kF 1
k cos
cos . Since and are two distinct acute angles, it follows that
kF 2
EA
EB
6=
kF 1
k , so that Leonardo was mistaken.
kF 2
15. The subspace consists of all vectors x in R 4 such that
2
4 x 1
3
2
4 1
3
xv =
x 2
x 3
x 4
5
2
3
4
5
= x 1 + 2x 2 + 3x 3 + 4x 4 = 0.
These are vectors of the form
2
4 2r3s4t
r
3
5
= r
2
4 2
1
0
0
3
5
+ s
2
4 3
3
5
+ t
2
4 4
0
0
1
3
5
.
s
t
The three vectors to the right form a basis.
235
k
EA
k
k
0
1
0
Chapter 5
ISM: Linear Algebra
16. You may be able to nd the solutions by educated guessing. Here is the systematic
approach: we rst nd all vectors x that are orthogonal to v 1 ; v 2 , and v 3 , then we identify
the unit vectors among them.
Finding the vectors x with xv 1 = xv 2 = xv 3 = 0 amounts to solving the system
x 1 + x 2 + x 3 + x 4 = 0
we can omit all the coecients
1
2
.
2
4 x 1
3
2
4 t t t
t
3
The solutions are of the form x =
x 2
x 3
x 4
5
=
5
.
Sincekxk= 2jtj, we have a unit vector if t =
1
2
or t = 2 . Thus there are two possible
choices for v 4 :
2
4 2 2
3
2
2 2
3
2
5
and
4
5
.
17. The orthogonal complement W ? of W consists of the vectors x in R 4 such that
2
4 x 1
3
2
4 1
3
2
4 x 1
3
2
4 5
3
x 2
x 3
x 4
5
2
3
4
5
= 0 and
x 2
x 3
x 4
5
6
7
8
5
= 0.
Finding these vectors amounts to solving the system
x 1 + 2x 2 + 3x 3 + 4x 4 = 0
5x 1 + 6x 2 + 7x 3 + 8x 4 = 0
.
The solutions are of the form
2
4 x 1
3
2
4 s + 2t
3
2
4 1 2
1
0
3
2
4 2 3
0
1
3
x 2
x 3
x 4
5
=
2s3t
s
t
5
= s
5
+ t
5
.
The two vectors to the right form a basis of W ? .
18. a.kxk 2 = 1+ 4 + 16 + 64 +=
1
1 4
= 3
use the formula for a geometric series, with a =
1
4
,
so thatkxk=
2 p
3
1:155.
236
x 1 + x 2 x 3 x 4 = 0
x 1 x 2 + x 3 x 4 = 0
ISM: Linear Algebra
Section 5.1
b. If we let u = (1; 0; 0; : : :) and v =
1; 2 ; 4 ;
, then
= arccos
kukkvk = arccos
uv
2 p
3
= arccos
p
3
2
= 6 (= 30 ).
does the job, since the harmonic series 1+ 2 + 3 +diverges
(a fact discussed in introductory calculus classes).
1;
2 ;
1 p
3 ;;
1 p
n ;
d. If we let v = (1; 0; 0; : : :); x =
1; 2 ; 4 ;
and u =
x kxk =
p
3
2
1; 2 ; 4 ;
then
proj L v = (uv)u =
3
4
1; 2 ; 4 ;
.
19. See Figure 5.2.
Figure 5.2: for Problem 5.1.19.
20. On the line L spanned by x we want to nd the vector mx closest to y (that is, we want
kmxykto be minimal). We want mxy to be perpendicular to L (that is, to x), which
means that x(mxy) = 0 or m(xx)xy = 0 or m =
xy
xx
4182:9
198:53 2
0:106.
Recall that the correlation coecient r is r =
xy
kxkkyk , so that m =
kyk
kxk r. See Figure 5.3.
21. Call the three given vectors v 1 ; v 2 , and v 3 . Since v 2 is required to be a unit vector, we
must have b = g = 0. Now v 1
v 2 = d must be zero, so that d = 0.
Likewise, v 2
v 3 = e must be zero, so that e = 0.
Since v 3 must be a unit vector, we havekv 3
k 2 = c 2 + 4 = 1, so that c = p
3
2
.
Since we are asked to nd just one solution, let us pick c =
3
2
.
237
c. x =
1 p
p
Chapter 5
ISM: Linear Algebra
Figure 5.3: for Problem 5.1.20.
The condition v 1
v 3 = 0 now implies that
p
3
2
a + 2 f = 0, or f = p
3a.
Finally, it is required thatkv 1 k 2 = a 2 + f 2 = a 2 + 3a 2 = 4a 2 = 1, so that a = 2 .
Let us pick a =
2 , so that f = p
3
2
.
Summary:
2
4 2
3
2
3
4 p
3
0
1
0
3
v 1 =
0
p
5
; v 2 =
4
5
; v 3 =
2
0
1
2
5
3
2
There are other solutions; some components will have dierent signs.
22. Let W =fx in R n : xv i = 0 for all i = 1; : : : ; mg. We are asked to show that V ? = W ,
that is, any x in V ? is in W , and vice versa.
, then xv = 0 for all v in V ; in particular, xv i = 0 for all i (since the v i
are in V ), so that x is in W .
?
, we have to verify that
xv = 0 for all v in V . Pick a particular v in V . Since the v i span V , we can write
?
238
1
2
If x is in V
Conversely, consider a vector x in W . To show that x is in V

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika:

Zgłoś jeśli naruszono regulamin