Page 208 - 35Linear Algebra

208 Linear Independence

This system has solutions if and only if the matrix M = v 1 v 2 v 3 is singular, so
we should find the determinant of M:
 
0 2 1
2 1
det M = det  0 2 4  = 2 det = 12.
2 4
2 1 3
Since the matrix M has non-zero determinant, the only solution to the system of
equations
1
 
c

c
v 1 v 2 v 3  2  = 0
c 3
is c 1 = c 2 = c 3 = 0. So the vectors v 1 , v 2 , v 3 are linearly independent.
Here is another example with bits:

3
Example 119 Let Z be the space of 3×1 bit-valued matrices (i.e., column vectors).
2
Is the following subset linearly independent?
     
 1 1 0 
1 , 0 , 1
     
0 1 1
 
If the set is linearly dependent, then we can find non-zero solutions to the system:

     
1 1 0
0
c 1   + c 2   + c 3   = 0,
1
1
0 1 1
which becomes the linear system
1
   
1 1 0 c
1 0 1 c = 0.
   2 
0 1 1 c 3
Solutions exist if and only if the determinant of the matrix is non-zero. But:
 
1 1 0
0 1 1 1
det  1 0 1  = 1 det − 1 det = −1 − 1 = 1 + 1 = 0
1 1 0 1
0 1 1
Therefore non-trivial solutions exist, and the set is not linearly independent.

Reading homework: problem 2

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