# Linear Subspace

**DEFINITION1:** Let be a field, be a -vector space and . If is also a -vector space, then is called a linear subspace or vector subspace (or shortly subspace) of .

**PROPOSITION1:** Let be a field, be a -vector space and . is a subspace of if and only if

**a)** ,

**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 ,

and

.

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:

The space

is the space of all bounded sequences. This space is a nontrivial subspace of .

to be continued...