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(MATH111)Ma111-midterm-06spring2.pdf
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Math111 L2: Linear Algebra, Midterm Test 2
Dept of Math, HKUST, Spring 2006 Name: ID No.:
Problem 1 (20) 2 (15) 3 (20) 4 (20) 5 (25) Total (100 pts)
Score

1. (5+7+8 pts) Find the bases for the vector spaces Nul A, Row A, and Col A, respectively, where
A =
. .
. . .
.
1 2 1 5 1 0
2 4 1 7 1 1
3 6 .1 3 .1 4
4 8 .2 2 .2 3

.
The vectors of the basis for Col A are required to be column vectors of A.
Solution.
. .
. . .
...
. . .
.

1 2 1 5 1 0
2 4 1 7 1 1
3 6 .1 3 .1 4
4 8 .2 2 .2 3

1 2 1 5 1 0
0 0 .1 .3 .1 1
0 0 .4 .12 .4 4
0 0 .6 .18 .6 3

. .
. . .
.
1 2 1 5 1 0
0 0 1 3 1 .1
0 0 0 0 0 0
0 0 0 0 0 .3


. .
. . .
.
1 2 0 2 0 0
0 0 1 3 1 0
0 0 0 0 0 1
0 0 0 0 0 0

Thus the basis for Nul A consists of vectors
v1 =[.2, 1, 0, 0, 0, 0]T , v2 =[.2, 0, .3, 1, 0, 0]T , v3 = [0, 0, .1, 0, 1, 0]T .

The following vectors
[1, 2, 0, 2, 0, 0]T , [0, 0, 1, 3, 1, 0]T , [0, 0, 0, 0, 0, 1]T .
The following columns vectors of A form a basis of Col A:
[1, 2, 3, 4]T , [1, 1 . 1, .2]T , [0, 1, 4, 3]T .

2. (15 pts) Let V be the subspace of R4 spanned by the vectors v1 = [1, 2, 3, 4]T , v2 = [2, 3, 4, 5]T . Find
a system of linear equations so that its solution space is V . Solution. Performing row operations:
. .
.

. .
.
12
x1
0 .1 x2 . 2x1
0 .2 x3 . 3x1
0 .3 x4 . 4x1


12
x1
0 1 2x1 . x2
0 0 x1 . 2x2 + x3
0 0 2x1 . 3x2 + x4

. .
.
12
x1
23
x2
x3
x4
Then V is the solution space of the linear system
x1 . 2x2 + x3 =0 2x1 . 3x2 + x4 =0
1
3. Let T : R4 R4 be de.ned by T
. .
. . .
.
. .
.
. .
.
=
. .
.
x1 + x2 + x3 + x4
2x1 +3x2 +4x3 +5x4
4x1 +5x2 +6x3 +7x4
3x1 +5x2 +7x3 +9x4
. .
.
. x1
x2
x3
x4
(a)
(7 pts) Determine whether T is 1-1. Explain brie.y why?

(b)
(7 pts) Determine whether T is onto. Explain brie.y why?

(c)
(6 pts) Find a basis for T (R4) := .T (x) | x R4..


Solution.
. .
. . .
. . .
. . .
. . .
. . .
.
1 1 1 1
0 1 2 3
0 0 0 0
0 0 0 0

1 1 1 1
2 3 4 5
4 5 6 7
3 5 7 9

1 1 1 1
0 1 2 3
0 1 2 3
0 2 4 6


(a)
T is not 1-1. This is because the standard matrix of T has only 2 pivot columns, not every column has a pivot position.

(b)
T is not onto. This is because not every row of the standard matrix of T has a pivot position.

(c)
The following vectors form a basis of T (R4).


[1, 2, 4, 3]T , [1, 3, 5, 5]T .
4. Let P3(t) be the vector space of polynomials of degree less or equal to 3.
(a)
(10 pts) Show that B = {t, t +1,t2 . 1,t3 +1} is a basis of P3(t).

(b)
(10 pts) Let D : P3(t) P3(t) be the derivative mapping, i.e.,


2
D.a0 + a1t + a2t2 + a3t3. = a1 +2a2t +3a3t.
Fin