Example: Take the vector space as the set of linear polynomials of degree 1.
A possible spanning set is
\[\{1,x \}\]
.With respect to this spanning set
\[2-x=2(1)+(-1)(x)\]
which we may write \[\begin{pmatrix}2\\-1\end{pmatrix}\]
.Another possible spanning set is
\[ \{1-x,1+x \}\]
.In terms of this spanning set
\[2-x=(1-x)+(1+x)-\frac{3}{2}(1-x) + \frac{1}{2}(1+x)\]
which we may write as the vector \[\begin{pmatrix}3/2\\1/2\end{pmatrix}\]
Every vector space has an associated dimension
\[n\]
and every set of elements has an order \[m\]
.
There are four possible cases.If
\[m the set of elements does not span the vector space.
If
There are four possible cases.
If
\[m=nn\]
the set of elements forms a basis for the vector space if and only if the set of elements of \[S\]
are linearly independent. If the set of elemtns of \[S\]
are linearly dependent then we can express at least one of the elements of \[S\]
in terms of other elemts, so throw out this element to get a small set with \[n>m\]
so \[S\]
is not a spanning set.There are four possible cases.
If
\[m>n\]
the set of elements is not linearly independent and may or may not span the vector space. We can reduce the siz of \[S\]
by expressing vectors in terms of other vectors, throwing out the vectors so expressed until we reduce \[S\]
to a linearly independent set with order \[s\]
.
If \[s the elemts of S do not span the vector space.
If
\[s=n\]
the elemts of S spans the vector space.
\[s>n\]
is not possible.