PROPOSITION1: Let be a field, be a -vector space and . is a subspace of if and only if
b) for all ,
c) for all and .
PROOF: (will be linked)
PROPOSITION2: Let be a field, be a -vector space and . is a subspace of if and only if for all and .
PROOF: (will be linked)
EXAMPLE1: Let be a field, be a -vector space. Then, there are always two subspaces of . They are and . Each of them is called a trivial subspace of . A subspace different from them is called nontrivial subspace.
EXAMPLE2: Let and be arbitrary constants. Then, the set is a subspace of . is different from the zero space for every . is nontrivial if and only if .
SOLUTION: Let and . Then, the equalities and are fulfilled.
Since , then . Moreover, the equality shows that the vector is in . These prove that is a subspace of .
Now, we prove the assertion for every . In the case , the relation holds because of the equality . On the other hand, in the case , the relation holds because of the equality . In either case, the subspace contains at least one vector different from the zero vector.
Now, we prove the proposition " is nontrivial if and only if ". We have already shown that is never equal to . So, it is sufficient to show that is different from if and only if . For this, we must prove two propositions: First: If , then . Second: If at least one of the numbers and is different from zero, then . The first proposition is obvious because for all if . We will prove the second proposition as . The case is similar. Let . Then, because the relation holds. Consequently, we conclude that .
EXAMPLE3: Consider the subset of . contains the zero vector because the equality is satisfied. Now, we show that is not a subspace of . Obviously, the vector is in . However, the scalar product is not in because . This example shows, in order that a subset is a subspace, it is not sufficient that the subset contains the zero vector. On the other hand, we know from Proposition1 that a subset not containing the zero vector is not a subspace.
EXAMPLE4: Let ,
We investigate whether each of and is a subspace of or not.
First, we investigate . For , we have .
Given and .
Then, the relation
shows that is a subspace.
Now, we investigate : in order that is a subspace, the zero vector must be in . For this reason, we must find two real numbers satisfying the equality . The equalities and give us . However, these values don't satisfy the equality . So, the zero vector is not in . This shows that is not a subspace.
EXAMPLE4 (SEQUENCE SPACES): In vector space-Example8, we have given all real sequences space . We will study some significant subspace of this space:
is the space of all bounded sequences. This space is a nontrivial subspace of .
to be continued...