Page 408 - 35Linear Algebra

408 Movie Scripts

1
Thus, for example, = 4 because 4 = 16 = 1 + 3 × 5 = 1. Don’t get too caught up
4
on this aspect, its a choice of base field designed to make computations go
quicker!
Now, here’s the problem we will solve:

   
1 0
3
2
   
Find a basis for V that includes the vectors   and   .
 3   2 
4 1
The way to proceed is to add a known (and preferably simple) basis to the
vectors given, thus we consider

           
1 0 1 0 0 0
3
0
0
0
1
2
           
v 1 =   , v 2 =   , e 1 =   , e 2 =   , e 3 =   , e 4 =   .
 3   2   0   0   1   0 
4 1 0 0 0 1
The last four vectors are clearly a basis (make sure you understand this....)
and are called the canonical basis. We want to keep v 1 and v 2 but find a way to
turf out two of the vectors in the canonical basis leaving us a basis of four
vectors. To do that, we have to study linear independence, or in other words
a linear system problem defined by
0 = α 1 e 1 + α 2 e 2 + α 3 v 1 + α 4 v 2 + α 5 e 3 + α 6 e 4 .
0
We want to find solutions for the α s which allow us to determine two of the
0
e s. For that we use an augmented matrix
 
1 0 1 0 0 0 0
 2 3 0 1 0 0 0 
  .
 3 2 0 0 1 0 0 
4 1 0 0 0 1 0
Next comes a bunch of row operations. Note that we have dropped the last column
of zeros since it has no information--you can fill in the row operations used
above the ∼’s as an exercise:
   
1 0 1 0 0 0 1 0 1 0 0 0
2 3 0 1 0 0 0 3 3 1 0 0
   
  ∼  
 3 2 0 0 1 0   0 2 2 0 1 0 
4 1 0 0 0 1 0 1 1 0 0 1
   
1 0 1 0 0 0 1 0 1 0 0 0
 0 1 1 2 0 0   0 1 1 2 0 0 
∼   ∼  
 0 2 2 0 1 0   0 0 0 1 1 0 
0 1 1 0 0 1 0 0 0 3 0 1

408