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MATH111 Quiz 3A Solutions
Page 1/4
November 4, 2008 by Daniel Zheng
..................................................40 mins allowed..................................................
Name: Student ID: Score:
..
.
I. [6] Let T : Rn Rm be a linear transformation with the standard matrix
.24 .2 .4 2 .6 .31 . .382 .3
..
.
Find a basis for the kernel and range of T respectively.
Solutions:
A basis for Ker T : ..
. .
. . .
. .6 .5 2 1 0 . .
. , . .
. .5 .3 2 0 1 . .
. ..
. .
. .
.2 4
..
. .
.
..
.
..
. ..
.
..
. ..
. .
.
A basis for Range T :2 , .6. .38
MATH111 Quiz 3A Solutions Page 2/4
II. Determine whether the following statements are true or not. If true, justify your answer; if false, give a counterexample.
1.
[2] A single vector by itself is linearly independent.
2.
[2] The nonpivot columns of a matrix are always linearly dependent.
3.
[2] If Nul A = {0}, then A must be invertible.
Solutions:
1. False. Consider {0}.
2. False.
3. False. A can be non-square.
MATH111 Quiz 3A Solutions Page 3/4
III. [4] Let B = {b1, , bn} and C = {c1, , cn} be bases for Rn . Show that the change-of-coordinates matrix P from B to C is invertible. (Hint: The j-th column of the change-of-coordinates matrix P is the coordinate vector [bj]C.) Proof. Let P =[p1, , pn], then pj =[bj]C. Suppose .ni=1 ipi = 0. Then
nnn0 =[c1, , cn] . ipi = .i[c1, , cn][bi]C = .ibi.
i=1 i=1 i=1 Since B is a basis, which should be linearly independent, we have i = 0 impliying {p1, , pn} are linearly independent. Hence P is invertible.
MATH111 Quiz 3A Solutions Page 4/4
IV. Let A be an m n matrix, and B be an n p matrix such that AB = 0. Show
1.
[3] Col B . Nul A.(. means subset to)
2.
[3] rankA + rankB n.(Hint: The Rank Theorem)
Proof.
1.
AB = A[b1, , bn]=[Ab1, ,Abn] = 0 implies the columns of B are solutions of Ax = 0 and hence belong to Nul A. Therefore, Col B = Span{b1, , bn}. Nul A.
2.
rankA + rankB = rankA + Dim ColB rankA + Dim NulA = n.