MATH 54 QUIZ 5 SOLUTIONS
FEB 26, 2014
GSI: MINSEON SHIN
(Last edited February 26, 2014 at 8:25pm.)
Problem 1. Let S be a subset of an n-dimensional vector space V , and suppose S contains fewer than n vectors. Explain
why S cannot span V .
Proof. Suppose that S spans V . By a theorem in the textbook, we can (if necessary) remove some vectors from S to obtain a
basis S 0 for V . Since S 0 is a basis of V which has less than n elements, this contradicts the assumption that V has dimension
n (since every basis of an n-dimensional vector space has n elements).
Problem 2. Assume that the matrix
bases for Col A, Row A, and Nul A.

1
2
A=
3
3
A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find
3 4
6 6
9 3
9 0
−1
0
6
9

2
−3

−3
0
,

1 3 4
0 0 1
B=
0 0 0
0 0 0
−1
−1
0
0

2
1

−5
0
Proof. The rank of A is the number of pivot columns, which is 3. Also, dim Nul A = 2 (the number of free variables). A
basis for Col A is (the pivot columns of A)
     
2 
4
1



     
−3
6
2
 , ,  ,
3 3 −3





0
0
3
a basis for Row A is (the nonzero rows of the REF of A)
{[1 , 3 , 4 , −1 , 2] , [0 , 0 , 1 , −1 , 1] , [0 , 0 , 0 , 0 , −5]} .
We describe the null space of A. If [x1 , x2 , x3 , x4 , x5 ] ∈ Nul A, then x5 = 0, x3 = x4 , and x1 + 3x2 + 4x3 − x4 + 2x5 = 0.
Thus every vector in the null space of A is of the form
  
 

 
x1
−3x2 − 3x4
−3
−3
x2  
0

1
x
2
  
 

 
x3  = 
 = x2  0  + x4  1  .
x4
  
 

 
x4  
1
0

x4
x5
0
0
0
Thus a basis for the null space of A is
   
−3
−3 





  0 

1

   
0 , 1 .
   


 0   1 






0
0
Problem 3. If A is a 7 × 5 matrix, what is the smallest possible dimension of Nul A?
Proof. The smallest possible dimension of Nul A is 0; for example, in the case


1 0 0 0 0
0 1 0 0 0


0 0 1 0 0



A=
0 0 0 1 0 .
0 0 0 0 1


0 0 0 0 0
0 0 0 0 0
(The answer is 2 if A is a 5 × 7 matrix, since the REF of A in this case will have at least 2 nonpivot columns.)
Problem 4. Let A be an m × n matrix. Which of the subspaces Row A, Col A, Nul A, Row AT , Col AT , Nul AT are in
Rm and which are in Rn ? How many distinct subspaces are in this list?
Proof. We have that Col A, Row AT , and Nul AT are subspaces of Rm , and Col AT , Row A, and Nul A are subspaces of Rn .
We always have Col A = Row AT and Col AT = Row A. Thus
there
are at most 4 different subspaces. An example in which
1 0 , in which case Col A = Row AT = R, Nul AT = {0}, and
there are exactly 4 different
subspaces
is
the
case
when
A
=
1
0
T
Col A = Row A = Span
, and Nul A = Span
.
0
1
Problem 5. Let
B=
b1 =
−1
1
, b2 =
8
−7
and
C=
c1 =
1
1
, c2 =
2
1
be bases for R2 . Find the change of coordinates matrix from B to C and the change of coordinates matrix from C to B.
Proof. The change of coordinates matrix from B to C is
P
C ← B = [b1 ]C
[b2 ]C
which you can compute by taking the right half of the REF of
−1 1 1 1
,
8 −7 2 1
which is
9
−10
−8
.
9
We can do a similar computation to show that
P
9
B←C =
10
8
.
9
1
Problem 6. Show that {ln 2, ln 3, ln 5} is linearly independent when R is considered as a vector space with scalars in Q.
a1 a2 a3
b1 , b2 , b3
such that
a1
a2
a3
ln 2 +
ln 3 +
ln 5 = 0
(1)
b1
b2
b3
where the ai , bi are integers and the bi are nonzero. We’d like to show that a1 = a2 = a3 = 0. Let m be a least common
multiple of b1 , b2 , b3 . Multiplying (1) by m gives
Proof. Suppose there exist rational numbers
c1 ln 2 + c2 ln 3 + c3 ln 5 = 0
where the ci =
mai
bi
(2)
are integers. Taking the exponential power of both sides in (2) gives
2c1 3c2 5c3 = 1 .
(3)
Depending on the signs of c1 , c2 , c3 , we have one of the following: 2|c1 | 3|c2 | 5|c3 | = 1, 2|c1 | 3|c2 | = 5|c3 | , 2|c1 | 5|c3 | = 3|c2 | ,
3|c2 | 5|c3 | = 2|c1 | , none of which can hold unless c1 = c2 = c3 = 0. Thus a1 = a2 = a3 = 0.
1Q is the set of rational numbers.