LINEAR ALGEBRA PROJECT 2. \/ECTOR. SPACES AND INNER PRODUCT SPACES tl (t) (u) Show that the set of vectors in iRa of the form {(r,2r,U,r*g)} forms a ' subspace and determine a basis for it. (b) Which...

1. (a) We wish to show that the vectors in 4R of the form  ,2 , ,x x y x y forms a subspace
W of 4R and to determine a basis for W.
We have

     
   
, 2 , , , 2 ,0, 0,0, ,
1, 2,0,1 0,0,1,1 .
x x y x y x x x y y
x y
  
 
Thus we see that
    1,2,0,1 0,0,1,1 | , ,W x y x y  R

whence     1,2,0,1 , 0,0,1,1 is a basis for W.

To see that W is a basis for 4R , we must show that it is closed under vector addition and
scalar multiplication. Let 1w and 2w be arbitrary vectors in W. Then there exist real
numbers 1 1 2 2, , , and x y x y such that
   1 1 11,2,0,1 0,0,1,1x y w and    2 2 21,2,0,1 0,0,1,1x y w .

Thus we have
       
     
1 2 1 1 2 2
1 2 1 2
1,2,0,1 0,0,1,1 1,2,0,1 0,0,1,1
1,2,0,1 0,0,1,1
.
x y x y
x x y y
W
    
   

w w
Thus we see that W is closed under vector addition.
Now let k be an arbitrary scalar. Then we have
   
   
1 1 1
1 1
1,2,0,1 0,0,1,1
1,2,0,1 0,0,1,1
.
k k x y
kx ky
W
   
 

w
Thus we see that W is also closed under scalar multiplication, whence it is a subspace of
4
R .
(b) We wish to determine which of the following sets are linearly independent over 4R .
(i)     1 1 26,3,7,0 , 4,6,2,1 .S v v  

Any set of two vectors is linearly dependent if and only if its vectors are
proportional. Since 1v and 2v are not proportional, we see that 1S is
linearly independent.
(ii)       2 1 2 33,7,2,1 , 4,0,1,6 , 7,7,3,7 .S u u u   

Note that 1 2 3,u u u  whence 2S is linearly dependent.
(c) (i) We wish to show that the vectors    1 26,7,1 , 2,0,3 ,v v  and  3 4,6,1v  form
a basis over 3.R
It suffices to show that 1 2 3, , and v v v are linearly independent over
3 ,R which
holds if and only if
1
2
3
6 7 1
2 0 3
4 6 1
v
A v
v
   
   
 
   
      
is nonsingular, which in turn holds if and only if det 0.A Now we have
6 7 1
det 2 0 3 .
4 6 1
A 
Expanding along the second row, we obtain
   
   
7 1 6 7
det 2 3
6 1 4 6
2 7 6 3 36 28
2 1 3 8
2 24
26.
A   
    
  
  
 
Since det 0,A  we see that A is nonsingular, whence 1 2 3, , and v v v are linearly
independent and hence form a basis over 3.R
(ii) We wish to find the coordinates of the vector  6,2,6v  with respect to
this basis.
We have 1 1 2 2 3 3v c v c v c v   for some constants 1 2 3, , and .c c c Thus we
have
       
     
 
1 2 3
1 1 1 2 2 3 3 3
1 2 3 1 3 1 2 3
6,2,6 6,7,1 2,0,3 4,6,1
6 ,7 , 2 ,0,3 4 ,6 ,
6 2 4 ,7 6 , 3 .
c c c
c c c c c c c c
c c c c c c c c
  
  
     
Equating components, we obtain the following linear system:
1 2 3
1 3
1 2 3
6 2 4 6;
7 6 2;
3 6.
c c c
c c
c c c
  
 
  
The corresponding augmented matrix is
6 2 4 6
7 0 6 2 .
1 3 1 6
 
 
 
  
Interchanging the first and third rows, we obtain
1 3 1 6
7 0 6 2 .
6 2 4 6
 
 
 
  
Subtracting seven times the first row from the second yields
1 3 1 6
0 21 1 40 .
6 2 4 6
 
 
  
 
  
Subtracting six times the first row from the third yields
1 3 1 6
0 21 1 40 .
0 16 2 30
 
 
  
 
    
Subtracting the third row from the second yields
1 3 1 6
0 5 1 10 .
0 16 2 30
 
 
 
 
    
Subtracting three times the second row from the third yields
1 3 1 6
0 5 1 10 .
0 1 5 0
 
 
 
 
   
Negating the third row yields
1 3 1 6
0 5 1 10 .
0 1 5 0
 
 
 
 
  
Interchanging the second and third rows, we obtain
1 3 1 6
0 1 5 0 .
0 5 1 10
 
 
 
   
Subtracting three times the second row from the first yields
1 0 14 6
0 1 5 0 .
0 5 1 10
 
 
 
   
Adding five times the second row to the third yields
1 0 14 6
0 1 5 0 .
0 0 26 10
 
 
 
  
Dividing the third row by 26 yields
5
13
1 0 14 6
0 1 5 0 .
0 0 1
 
 
 
  
Adding 14 times the third row to the first yields
8
13
5
13
1 0 0
0 1 5 0 .
0 0 1
 
 
 
  
Finally, subtracting five times the third row from the second yields
8
13
25
13
5
13
1 0 0
0 1 0 .
0 0 1
 
 
 
  
Thus we have 8 25 51 2 313 13 13, , and ,c c c    whence
       8 25 513 13 136,2,6 6,7,1 2,0,3 4,6,1 .  
(iii) We wish to compute the transition matrix C BP  from the standard basis
 1 2 3, ,B e e e to the basis  1 2 3, , .C v v v

We have
     1 2 3
6 2 4
7 0 6 .
1 3 1
C B B B B
P v v v    
 
 

 
  
(d) We wish to determine the dimension of the subspace W of 4R spanned by the vectors
 1 3,6,3,0v  ,  2 4,2,1,1v  , and  3 2,0,2, 2v   , and to find a basis for W and give a
general form for a vector in W.
To determine the dimension of W, we apply row reduction to the matrix
1
2
3
3 6 3 0
4 2 1 1 .
2 0 2 2
v
A v
v
   
   
 
   
      
Interchanging the first and third rows, we obtain
2 0 2 2
4 2 1 1 .
3 6 3 0
 
 
 
  
Subtracting twice the first row from the second yields
2 0 2 2
0 2 3 5 .
3 6 3 0
 
 

 
  
Dividing the first row by 2 yields
1 0 1 1
0 2 3 5 .
3 6 3 0
 
 

 
  
Subtracting three times the first row from the third yields
1 0 1 1
0 2 3 5 .
0 6 0 3
 
 

 
  
Dividing the third row by 2 yields
1 0 1 1
0 2 3 5 .
0 2 0 1
 
 

 
  
Interchanging the second and third rows, we obtain
1 0 1 1
0 2 0 1 .
0 2 3 5
 
 
 
  
Subtracting the second row from the third yields
1 0 1 1
0 2 0 1 .
0 0 3 4
 
 
 
  
Dividing the second row by 2 yields
1
2
1 0 1 1
0 1 0 .
0 0 3 4
 
 
 
  
Dividing the third row by...