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# MTH501 assignment no 3 (due date 8 January 2017)

For a nonempty subset ${\displaystyle S}$ of a vector space, under the inherited operations, the following are equivalent statements
1. ${\displaystyle S}$ is a subspace of that vector space
2. ${\displaystyle S}$ is closed under linear combinations of pairs of vectors: for any vectors ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$ and scalars ${\displaystyle r_{1},r_{2}}$ the vector ${\displaystyle r_{1}{\vec {s}}_{1}+r_{2}{\vec {s}}_{2}}$ is in ${\displaystyle S}$
3. ${\displaystyle S}$ is closed under linear combinations of any number of vectors: for any vectors ${\displaystyle {\vec {s}}_{1},\ldots ,{\vec {s}}_{n}\in S}$ and scalars ${\displaystyle r_{1},\ldots ,r_{n}}$ the vector ${\displaystyle r_{1}{\vec {s}}_{1}+\cdots +r_{n}{\vec {s}}_{n}}$ is in ${\displaystyle S}$.
Anyone have idea about it, please leave comment and  am trying  to solve

S = {
 0 4 1
,
 1 2 3
}
Of vectors in the vector space R3, find a basis for span S.
The set S = {v1v2} of vectors in R3 is linearly independent if the only solution of
c1v1 + c2v2 = 0
is c1, c2 = 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 v1v2, (*) becomes:
c1
 0 4 1
+
c2
 1 2 3
=
 0 0 0
Rearranging the left hand side yields
 0 c1 +1 c2 4 c1 +2 c2 1 c1 +3 c2
=
 0 0 0
The matrix equation above is equivalent to the following homogeneous system of equations

 0 c1 +1 c2 = 0 4 c1 +2 c2 = 0 1 c1 +3 c2 = 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).
 0 1 4 2 1 3
can be transformed by a sequence of elementary row operations to the matrix
 1 0 0 1 0 0

The reduced row echelon form of the coefficient matrix of the homogeneous system is
 1 0 0 1 0 0
which corresponds to the system
 1 c1 = 0 1 c2 = 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 c1, c2 = 0.
Therefore the set S = {v1v2} is linearly independent.

Consequently, the set S forms a basis for span S and many set of solution .