Subspace proof.

Question

If U and W are subspaces of a vector space V, prove that U intersect W is also a subspace of V. Is the union of V and W a subspace of V?

Answer 1

Before we begin this proof, I want to make sure we are clear on the definition of a subspace.

Let V be a vector space over a field K. W is a subspace of V if it satisfies the following properties...

1. W is a non-empty subset of V
2. If w₁ and w₂ are elements of W, then w₁+w₂ is also an element of W (closure under addition)
3. If c is an element of K and w is an element of W, then cw∩ is also an element of W (closure under scalar multiplication)

To prove that U intersection with W is a subspace, we need to show the above three properties are satisfied. Now let's begin our proof...

Let S=U∩W.

Property 1:

U and W are both subspaces of V thus U and W are both subsets of V (U,W⊆V)

The intersection of two sets will contain all members of the two sets that are shared. This implies S ⊆ V.

Since both U and W contain 0 (as is required for all subspaces), S also contains 0 (0∈S). This implies that S is a non empty subset of V.

Property 2:

Now we must show closure under addition...

Let s₁ and s₂ be elements of S. We need to show that s₁+s₂ ∈ S.

Since s₁,s₂∈S, this implies that s₁,s₂ ∈ U,W.

Since s₁,s₂ ∈ U, then s₁+s₂ ∈ U (U is a vector space).

Similarly, Since s₁,s₂ ∈ W, then s₁+s₂ ∈ W (W is a vector space).

Since s₁+s₂ ∈ U,W, it follows that s₁+s₂ ∈U∩W=S

This implies S is closed under addition.

Property 3:

Finally, we must show closure under scalar multiplication...

Let s ∈ S and let c be a scalar.

We must show cs ∈ S.

Given s ∈ S, this implies s∈U,W.

Since U and W are vector spaces, this implies cs ∈ U,W.

Since cs∈U,W, it follows that cs ∈ U∩W=S.

This implies S is closed under scalar multiplication.

All three properties are satisfied, therefore S is a subspace.

Thus completes our proof

Now to address whether the union of two subspaces are a subspace...

I will provide a counterexample...

It can be shown that U=span{(1,0) and W=span{(0,1)} will are both subspaces of R^T.

It can also easily be show that (1,0) ∈ U and (0,1) ∈ W.

Now, let S=U ∪ W

It follows that (1,0) and (0,1) are both elements of S.

Now, I will show property 2 fails...

(1,0)+(0,1)=(1,1)

(1,1) ∉ U and (1,1) ∉ W, therefore (1,1) ∉ U∪W= S.

Thus, U ∪ W will not form a subspace of V in general.