# Intersection of Two Subspaces is a Subspace (Step by Step Proof)

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In this article, we give a step by step proof of the fact that the intersection of two vector subspaces is also a subspace.

The proof is given in three steps which are the following:

1. The zero vector lies in the intersection of the subspaces.
2. The intersection is closed under the addition of vectors.
3. The intersection is closed under multiplication by scalars.

### Proof:

Let W be a vector space and U and V be two subspaces of the vector space. Then, U∩V is also a vector subspace.

#### Step 1: Show that 0 ∈ U∩V

Since U itself is a vector space it contains the zero vector, which is

0 ∈ U (EQUATION 1)

Similarly, since V is also a vector space we get,

0 ∈ V (EQUATION 2)

Combining (EQUATION 1) and (EQUATION 2) we see that,

0 ∈ U∩V

This proves that the intersection contains the zero vector and is non-empty.

#### Step 2: Show that if x ∈ U∩V and y ∈ U∩V, then x + y ∈ U∩V

Since x ∈ U∩V, this implies that x ∈ U. Similarly y ∈ U∩V, this implies that y ∈ U. Now, since U is a vector space it is closed under vector addition.

x ∈ U and y ∈ U ⇒ x + y ∈ U (EQUATION 3)

Similarly, since V is closed under vector addition we get,

x + y ∈ V (EQUATION 4)

Combining (EQUATION 3) and (EQUATION 4) we conclude that,

x + y ∈ U∩V

This proves that U∩V is closed under vector additon.

#### Step 3: Show that if x ∈ U∩V and α is a scalar, then α.x ∈ U∩V

Since x ∈ U∩V, this implies that x ∈ U. Since U is a vector space it is closed under scalar multiplication. This means that,

α.x ∈ U (EQUATION 5)

Similarly, x also belongs to the vector space V which is also closed under scalar multiplication. We have,

α.x ∈ V (EQUATION 6)

Combining (EQUATION 5) and (EQUATION 6) we get,

α.x ∈ U∩V

This proves that U∩V is closed under scalar multiplication.

The above three steps complete the proof of the theorem. We summarize the proof strategy in the image below.