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APSC 174J, Solutions #8
Posted: May 19, 2016
Section 11, Problem 1(b).
The linear transformation determined by A is LA : R → R̂2 , defined by
 
−x
LA (x) =  
2x
 
 
−1
−1
(1) The Range Im(A) is the linear span of { }. Since   is a non-zero vector,
2
2
it is linearly independent, hence a basis for Im(A). Therefore, dim(Im(A))=1.
(2) To find the Kernel of A, we solve for LA (x) = 0:
   
−x
0
 = 
2x
0
the solution is x = 0. Hence Ker(A)={0}, and dim(Ker(A))=0.
(3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=0. The rank-nullity theorem
is satisfied:
1 + 0 = 1 = the number of columns of A
Section 11, Problem 1(d).
The linear transformation determined by A is LA : R̂3 → R̂2 , defined by
 


x1
 
3x + 2x2 + x3
) =  1

LA (
x
2
 
0
x3
     
3
2
1
(1) The Range Im(A) is the linear span of {  ,   ,  }. Easy to see that
0
0
0
 
 
 
 
 
3
1
2
1
1
  = 3   and   = 2  , hence a basis for Im(A) is { }. Therefore,
0
0
0
0
0
dim(Im(A))=1.
2
(2) To find the Kernel of A, we solve for LA (x) = 0:
  

0
3x + 2x2 + x3
= 
 1
0
0
the solution is x3 = −3x1 − 2x2 and x1 , x2 can be any real numbers. Write the
solution in parametric vector form we have
  

 
 
x1
x1
1
0
  

 
 
 
 = x1  0  + x2  1 
x=
x2
x2  = 

 
 
x3
−3x1 − 2x2
−3
−2

1
 
0


1
 
0

   
   
 ,  1 }. Check that { 0  ,  1 } is
So Ker(A) is the linear span of {
0
   
   
−3
−2
−3
−2
   
0
1
   
  
linearly independent. Hence (
 0  ,  1 ) is a basis for Ker(A), and
−2
−3
dim(Ker(A))=2.
(3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem
is satisfied:
1 + 2 = 3 = the number of columns of A
Section 11, Problem 1(e).
The linear transformation determined by A is LA : R̂2 → R̂2 , defined by
 
 
x1
0
LA ( ) =  
x2
0
so LA is the zero transformation.
   
0
0
(1) The Range Im(A) is the linear span of {  ,  }, i.e. Im(A)= {0} Therefore,
0
0
dim(Im(A))=0.
(2) To find the Kernel of A, we solve for LA (x) = 0. Since LA (x) = 0 for any x ∈ R̂2 ,
we have Ker(A)= R̂2 , and dim(Ker(A))=2.
3
(3) Rank(A)=dim(Im(A))=0, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem
is satisfied:
0 + 2 = 2 = the number of columns of A
Section 11, Problem 1(r).
The linear transformation determined by A is LA : R̂4 → R̂2 , defined by
 
x1


 
x 
3x2
 2

LA ( ) = 
x3 
−x
3
 
x4
       
0
3
0
0
(1) The Range Im(A) is the linear span of {  ,   ,   ,   , }. Easy to see
0
0
−1
0
   
3
0
that a basis for Im(A) is {  ,   , }. Therefore, dim(Im(A))=2.
0
−1
(2) To find the Kernel of A, we solve for LA (x) = 0:
 
0
3x
 2 =  
0
−x3


the solution is x2 = x3 = 0 and x1 , x4 can be any real numbers. Write the
solution in parametric vector form we have
   
 
 
x1
x1
1
0
   
 
 
x   0 
0
0
 2  
 
 
x =   =   = x1   + x4  
x3   0 
0
0
   
 
 
x4
x4
0
1
   
   
1
0
1
0
   
   
0 0
0 0
   
   
So Ker(A) is the linear span of {  ,  }. Check that {  ,  } is linearly
0 0
0 0
   
   
0
1
0
1
4
   
1
0
   
0 0
   
independent. Hence (  ,  ) is a basis for Ker(A), and dim(Ker(A))=2.
0 0
   
0
1
(3) Rank(A)=dim(Im(A))=1, Nullity(A)=dim(Ker(A))=2. The rank-nullity theorem
is satisfied:
2 + 2 = 4 = the number of columns of A
Section 11, Problem 1(z).
The linear transformation determined by A is LA : R̂4 → R̂3 , defined by
 


x1
 
x
+
x
+
2x
−
x
1
2
3
4
x 


 2

LA ( ) = 
3x
+
6x
+
3x
1
3
4


x3 
 
−2x1 − 4x3 + 2x2
x4
      
−1
2
1
       
      
(1) The Range Im(A) is the linear span of {
 3  , 0 ,  6  ,  3 }. To find a
2
−4
0
−2
basis of Im(A), we reduce A to row-echelon form:


1





−1
1 1 2 −1
1 1 2 −1






3 0 6
 → 0 −3 0 6  → 0 −3 0 6 
3






−2 0 −4 2
0 2 0 0
0 0 0 4
1
1
2
Columns with leading
arethe first, second and fourth column. Thus a
 entries
  
1
1
−1
     
    
basis for Im(A) is {
 3  , 0 ,  3 }. Therefore, dim(Im(A))=3.
2
0
−2
(2) To find the Kernel of A, we solve for LA (x) = 0:

  
x1 + x2 + 2x3 − x4
0

  
 3x1 + 6x3 + 3x4  = 0

  
−2x1 − 4x3 + 2x2
0
5

1
1
2

 3 0 6

−2 0 −4
−1 0
3
2


1
1
2 −1 0


 0 −3 0
→
0 


0
0 0 0
6
4


0 

0
the solution is x4 = 0, x2 = 2x4 = 0, x1 = −2x3 and x3 is free. Write the solution
in parametric vector form we have
  

 
x1
−2x3
−2
  

 
x   0 
0
 2 

 
x= =
 = x3  
x3   x3 
1
  

 
x4
0
0
 
 
−2
−2
 
 
0
0
 
 
So Ker(A) is the linear span of { }. Since   is a non-zero vector, it is
1
1
 
 
0
0
 
−2
 
0
 
linearly independent. Hence ( ) is a basis for Ker(A), and dim(Ker(A))=1.
1
 
0
(3) Rank(A)=dim(Im(A))=3, Nullity(A)=dim(Ker(A))=1. The rank-nullity theorem
is satisfied:
3 + 1 = 4 = the number of columns of A