## Linear Independence

A set of vectors
$S$
is linearly indpendent of none of the vectors in
$S$
can be expressed in terms of the others. Any set
$S$
of elements (as long as
$S$
contains more than one element) may or may not be linearly indpendent.
One way to determine independence to to write the elements of
$S$
as column vectors with respect to some basis. If the resulting matrix is square (
$n$
vectors in a space of dimension
$n$
) then the vectors are linearly independent if the determinmant of the matrix is not zero.
Example:
$S = \{1,1+x,1+x+x^2 \}$
With respect to the standard basis
$B= \{1,x,x^2 \}$
. We can write the elements of
$S$
as
$\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}$
.
The resulting matrix is
$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$
with determinant 1 and the elements of
$S$
are linearly independent. If
$S = \{1+x,1+x^2,2+x+x^2 \}$
With respect to the standard basis
$B= \{1,x,x^2 \}$
then we can write the elements of
$S$
as
$\begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}2\\1\\1\end{pmatrix}$
.
The resulting matrix is
$\left( \begin{array}{ccc} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array} \right)$
with determinant 0 and the elements of S are linearly dependent.
It should be noted that linear dependence or independence is a property of the set and not the basis. Chaning the basis does not affect linear dependence.
Example:
$S = \{1+\mathbf{i},1, 1-\mathbf{i}\}$
With respect to the standard basis
$B= \{1, \mathbf{i} \}$
we can write the elements of
$S$
as
$\begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\-1\end{pmatrix}$
.
The resulting matrix is
$\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -1 \end{array} \right)$
.
This is not a square matrix so has no determinant. The dimesion of the set of complex numbers is 2, and there are three elemtns in this set. The order of
$S$
is greater than the dimension of the space so
$S$
is linearly dependent.