Subspace Proof

The concept of linear subspace or vector subspace as it is sometimes called is very important in linear algebra and various other fields of physics and math. A linear subspace is commonly referred to as subspace when the context serves to distinguish it from other kind of subspaces. The subspace proof explained.

Let K be the field and V be the vector space over K. Let the element of V be vectors and that of K be scalars. Now let W be the subset of V. If W is itself the vector space with the same vector space operations as V has then it is known as a subspace of V.

In order to use this definition and as a subspace proof we need not prove that all the propertites of vector space hold for W. Instead of it, we can prove a theorem that will give us an easier way to show that a subset of vector space is a subspace.

Assume that V is the vector space over the field K and also assume that W is the subset of V. Then we can say that W is a subset when

- The zero vector is in W.
- Let U and V are the elements of W, then the sum U+V is an element of W.
- If C is a scalar from K and U is an element of W, then the scalar product, i.e. CU is an element of W.

Proof
The first property confirms that W is non- empty. Also the second and third property assumes that closure of W under the addition and the scalar multiplication; therefore the vector space orientations are well defined.

The elements of W are also elements of V, so axioms 1-2 and 5-8 of the vector space are satisfied. By closing W under scalar multiplication the axioms 3 and 4 of a vector space are satisfied.

Conversely if W is a subspace of V. Then W itself will become a vector space under the various operations induced by V. As a result of it the second and third properties are also satisfied. By the virtue of property third, -w is in W wherever w is. And it also follows that W is closed under subtraction as well. Now since W is non empty therefore there is an element x in W, and x-x=0 is also in W. Therefore property one is also satisfied.

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