MTH501 assignment no 3 (due date 8 January 2017)
For a nonempty subset
$S$ of a vector space, under the inherited operations, the following are equivalent statements
 $S$ is a subspace of that vector space
 $S$ is closed under linear combinations of pairs of vectors: for any vectors ${\vec {s}}_{1},{\vec {s}}_{2}\in S$ and scalars $r_{1},r_{2}$ the vector $r_{1}{\vec {s}}_{1}+r_{2}{\vec {s}}_{2}$ is in $S$
 $S$ is closed under linear combinations of any number of vectors: for any vectors ${\vec {s}}_{1},\ldots ,{\vec {s}}_{n}\in S$ and scalars $r_{1},\ldots ,r_{n}$ the vector $r_{1}{\vec {s}}_{1}+\cdots +r_{n}{\vec {s}}_{n}$ is in $S$.
Anyone have idea about it, please leave comment and am trying to solve
Idea About Problem no 2
Of vectors in the vector space R^{3}, find a basis for span S.
The set S = {v_{1}, v_{2}} of vectors in R^{3} is linearly independent if the only solution of
c_{1}v_{1} + c_{2}v_{2} = 0
is c_{1}, c_{2} = 0.
In this case, the set S forms a basis for span S.
Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent.
If this is the case, a subset of S can be found that forms a basis for span S.
With our vectors v_{1}, v_{2}, (*) becomes:
Rearranging the left hand side yields
0 c_{1} +1 c_{2}

4 c_{1} +2 c_{2}

1 c_{1} +3 c_{2}



=


The matrix equation above is equivalent to the following homogeneous system of equations

0 c_{1}

+1 c_{2}

=

0

4 c_{1}

+2 c_{2}

=

0

1 c_{1}

+3 c_{2}

=

0







We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has
 the trivial solution only (meaning that S is linearly independent), or
 the trivial solution as well as nontrivial ones (S is linearly dependent).
can be transformed by a sequence of elementary row operations to the matrix
The reduced row echelon form of the coefficient matrix of the homogeneous system is
which corresponds to the system
1 c_{1}


=

0


1 c_{2}

=

0


0

=

0

Since each column contains a leading entry (highlighted in yellow), then the system has only the trivial solution, so that the only solution of is c_{1}, c_{2} = 0.
Therefore the set S = {v_{1}, v_{2}} is linearly independent.
Consequently, the set S forms a basis for span S and many set of solution .