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MTH501 assignment no 3

MTH501 assignment no 3 (due date 8 January 2017)




For a nonempty subset  of a vector space, under the inherited operations, the following are equivalent statements
  1.  is a subspace of that vector space
  2.  is closed under linear combinations of pairs of vectors: for any vectors  and scalars  the vector  is in 
  3.  is closed under linear combinations of any number of vectors: for any vectors  and scalars  the vector  is in .
Anyone have idea about it, please leave comment and  am trying  to solve

Idea About Problem no 2
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 .