MTH501 assignment no 3

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

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 = {v₁, v₂} 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 v₁, v₂, (*) 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 = {v₁, v₂} is linearly independent.

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