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.


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


  NullSpace[a]

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:


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


  MatrixForm[m]


  NullSpace[m]

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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