MATH2040 HW1 Solution
Tsang Chi Shing, Sidney and Lo Chiu Hong
Due Date: 19-1-2012
Sec. 5.1: 3(c), 3(d), 4(g), 4(h) 6, 7
3(c)
(i) The characteristic polynomial of A is:
it
1
2
i t
det(A tI ) =
= (t + 1)(t 1)
Therefore the eigenvalues of A are 1 and 1.
(ii
MATH2040 HW2 Solution
Tsang Chi Shing and Yu Rongfeng
Sec. 5.1: 8, 9, 10, 11, 13, 15, 17, 20, 21, 22; Sec. 5.2: 2(b), 2(c), 2(e).
8 (a) Let T be a linear operator on a nite-dimensional vector space V .
T is invertible,
N (T ) = cfw_0 (N (T ) is the null
MATH2040 HW 4 Solution
Tsang Chi Shing, Yu Rongfeng
Sec. 5.2: 17, 18, 19, 20, 21, 23
17) (a) Suppose that T and U are simultaneously diagonalizable linear operators on a nite-dimensional vector space V , then there exists an
ordered basis for V such that
MATH2040 HW 10 Solution
Tsang Chi Shing, Yu Rongfeng
Sec. 6.6: 4, 5, 6, 7, 9, 10
4) Let W be a nite-dimensional subspace of an inner product space V .
Suppose that T is the orthogonal projection of V on W , then for any
y V , T (y ) = u for y = u + z with
MAT2040 Assignment 11 Solution
Tsang Chi Shing, Sidney and Lo Chiu Hong
Due Date: 5-4-2012
Sec. 6.8: 5(a)(c), 7, 8, 13, 16, 22(a)(b)
5 (a) H : R3 R3 R, where
a1
b1
H a2 , b2 = a1 b1 2a1 b2 + a2 b1 a3 b3
a3
b3
z1
x1
y1
For all x = x2 , y = y2 , z = z2 R3 a
MATH2040 HW 12 Solution
Tsang Chi Shing, Yu Rongfeng
Sec. 7.1: 2, 4, 5, 8, 10; Sec. 7.2: 2, 3, 4, 6, 7, 11, 12, 13, 14, 15, 16, 17.
Sec. 7.1:
2) (a) The characteristic polynomial of A is:
1t
1
det(A tI ) = det
1
3t
= (t 2)2
Hence = 2 is the only eigenvalu
MATH2040 Linear Algebra II Midterm 2
Name:
Student ID:
You have to answer all seven questions. The total score is 110.
Please show your steps unless otherwise stated. No calculators are allowed.
1. Let
3
1
A=
1
1
1
3
1
1
1
1
3
1
1
1
.
1
3
(a) (10pts) Find