Linear Independence

A set of vectors {F1, F2, . . . , Fn} is linearly independent if and only if
a1F1+a2F2+ . . . + anFn=0 implies a1=a2=. . .=an=0.
To test if the vectors (1,0,3,0), (0,1,-6,1), (0,2,1,-1) are linearly independent in 4 dimensional space, consider
c1(1,0,3,0)+c2(0,1,-6,1)+c3(0,2,1,-1)=(0,0,0,0). This can be written as
1c1+0c2+0c3=0
0c1+1c2+2c3=0
3c1-6c2+1c3=0
0c1+1c2-1c3=0
If you had linear algebra this is old hat, but if not, place the vectors in the columns of a matrix A and solve Ax=0.

In[59]:=

  a=Transpose[{{1,0,3,0},{0,1,-6,1},{0,2,1,-1}}];
  MatrixForm[a]

Out[59]=

  1    0    0
  
  0    1    2
  
  3    -6   1
  
  0    1    -1

In[60]:=

  NullSpace[a]

Out[60]=

{}

There is no solution (other than the trivial one (0,0,0) ) so the vectors are linearly independent.

To see if the vectors (1,-9), (4,5), and (2,-6) are linearly independent in R2:

In[61]:=

  m=Transpose[{{1,-9},{4,5},{2,-6}}];

In[62]:=

  MatrixForm[m]

Out[62]=

  1    4    2
  
  -9   5    -6

In[63]:=

  NullSpace[m]

Out[63]=

  {{-34, -12, 41}}

They are not! In fact, -34(1,-9)-12(4,5)+41(2,-6)=(0,0). If you had linear algebra, you knew already that the THREE vectors could not be linearly independent in TWO dimensional space.

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