Handout - Denition of a Vector Space
Math 1185 Honors Linear Algebra - RUBIN - September 24, 2004
Consider a set V on which the operations addition (denoted by +) and scalar multiplication are dened. Together, these form a vector space if V is closed unde
Weeks 10-11
Math 1185 Honors Linear Algebra - RUBIN - Fall 04
SCHEDULE: Bold-faced homework problems from Sections 5.1, 5.2, and 5.3 will be due in
class on Monday, November 8th. Bold-faced homework problems from Section 5.5 will be
due in class on Monday
Weeks 12-13
Math 1185 Honors Linear Algebra - RUBIN - Fall 04
SCHEDULE: Bold-faced homework problems from Sections 5.6 and from 6.1 where specied will be due in class on Monday, November 22nd. Bold-faced homework problems from
Section 6.1 where specied an
Solutions to Homework for Week 1 - Math 1185 - Fall, 2004 - Rubin
1.1, #3. (a) Think of these equations as x2 = x1 + 4, x2 = x1 2. These 2 lines
intersect in a point (the rst has negative slope in (x1 , x2 ) plane, the second has
positive slope), so there
Solutions to Homework for Week 2 (Section 1.3) - Math 1185 - Fall, 2004 - Rubin
1.3, #13. (a) b = 2a1 + a2 . (b) Clearly one solution is (x1 , x2 ) = (2, 1). Since there are
2 independent equations with 2 unknowns, the solution is unique. (c) Rather than
Solutions to Homework for Week 3 (Sections 1.4-5) - Math 1185 - Fall, 2004 - Rubin
10
1.4, #5. a) E = 0 1
10
F EA = C for F, E
0
10 0
0 . b) F = 0 1 1 . c) C is row-equivalent to A, because
1
00 1
both elementary.
1.4, #6. a) To transform A to upper trian
Solutions to Homework for Week 4 - Math 1185 - Fall, 2004 - Rubin
2.1, #3. (c) Since rows 2 and 3 are identical, the determinant is 0. (d) The determinant
is 4(1)(4) + (3)(2)(5) (3)(3)(4) (4)(2)(1) = 58. (g) By cofactor expansion
on the second row, the de
Solutions to Homework for Weeks 5-6 - Math 1185 - Fall, 2004 - Rubin
3.2, 8. a) Suppose B1 , B2 S . Then A(B1 + B2 ) = AB1 + AB2 = B1 A + B2 A = (B1 +
B2 )A, so B1 + B2 S . Further, for scalar c, A(cB1 ) = c(AB1 ) = c(B1 A) = (cB1 )A,
so cB1 S as well and
Solutions to Homework for Week 7 - Math 1185 - Fall, 2004 - Rubin
3.5, 1a,2a. See back for 1a; invert 1a to get 2a.
1/2 1/2
1/2 1/2
3.5, 3a. SV U = SeU SV e =
34
23
=
5/2
7/2
1/2 1/2
1
3.5, 4. [z ]E = SeE [z ]e . We have [z ]e = (10, 7)T and SeE = SE e =
Solutions to Homework for Week 10 - Math 1185 - Fall, 2004 - Rubin
T
v
projection of v onto w is p = wT w w =
w
lies on w already so its projection onto w is itself.
5.1, 2b) =
0.
0
13
vT w
42 =
|w|2 =
126
42
1
126 w = 3 w = (2, 1, 3).
5.1, 2a) The scalar
Solutions to Homework for Week 11 - Math 1185 - Fall, 2004 - Rubin
5.5, 2a) Evaluate uT u2 uT u3 uT u3 = 0 and uT ui = 1 for i = 1, 2, 3.
1
1
2
i
5.5, 2b) Our goal is to nd c1 , c2 , c3 such that x = (1, 1, 1)T = c1 u1 + c2 u2 + c3 u3 .
Theorem 5.5.2 impl
Solutions to Homework for Week 12 - Math 1185 - Fall, 2004 - Rubin
5.6, 1. a) Since the columns of A are linearly independent, R(A) is 2-dimensional and
cfw_(1, 0)T , (0, 1)T forms an orthonormal basis for R(A). Using the Gram-Schmidt pro
1
1
cess instea
Solutions to Homework for Week 14 - Math 1185 - Fall, 2004 - Rubin
ASPG, 1. Consider
1
P Dk P 1 x(0)
k
1
1
0
.
0
2 k
0 ( )
.
0
1
=P
.
0
0
. . . ( n )k
1
1 (0)
P x .
As k , since |j | < 1 for all j = 1, this expression converges to
1 0 . 0
0 . 0
c
0 0