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:

- The zero vector lies in the intersection of the subspaces.
- The intersection is closed under the addition of vectors.
- 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.