# Sum of Vector Subspaces

Given a vector space V over the field K, and two subspaces A and B of V, the sum A+B of the subspaces is a vector subspace formed by the sum of vectors from A and B. $$ A+B := \{ \vec{a}+\vec{b}, \vec{a} \in A, \vec{b} \in B \} $$ Thus, the sum of the subspaces encompasses the union of the two subspaces $$ A∪B \subset A+B $$ Moreover, A+B is the smallest vector subspace containing the union of the subspaces A∪B.

The sum of two subspaces A+B is a subspace of the vector space V.

It is referred to as the **sum subset** of V.

$$ A+B \in V $$

Being a vector space, V inherits the sum properties of vector spaces.

In turn, the sum of the subspaces A+B includes both subspaces A and B.

$$ A \in A+B $$

$$ B \in A+B $$

Furthermore, the space A+B is **the smallest vector subspace of V that contains both A and B** as subspaces.

**Note**. The union A∪B is not a vector subspace because the union of subspaces is not itself a vector subspace.

## Proof

__The sum of the subspaces A+B is a vector subspace__

First, I need to demonstrate that the sum of vector subspaces is a vector subspace.

It's important to note that the subspaces A and B include the zero vector 0.

Therefore, the set A+B also contains the **zero vector**.

$$ \vec{a} + \vec{b} = \vec{0} + \vec{0} = \vec{0} \in A+B $$

Now I can verify if it meets the two properties of vector subspaces.

**1) Sum Property**

Let's take two generic vectors u and w from A+B.

$$ \vec{u},\vec{w} \in A+B $$

Where u and w are vectors obtained by summing a vector from A with a vector from B

$$ \vec{u} = \vec{a_1} + \vec{b_1} $$

$$ \vec{w} = \vec{a_2} + \vec{b_2} $$

with

$$ \vec{a_1} \ , \ \vec{a_2} \in A $$

$$ \vec{b_1} \ , \ \vec{b_2} \in B $$

The sum of the vectors

$$ \vec{u} + \vec{w} = (\vec{a_1} + \vec{b_1} ) + (\vec{a_2} +\vec{b_2} ) $$

Can be rewritten, thanks to the associative property, as

$$ \vec{u} + \vec{w} = (\vec{a_1} + \vec{a_2} ) + ( \vec{b_1} + \vec{b_2} ) $$

The sum of the vectors a_{1}+a_{2} is a sum vector belonging to the vector subspace A.

Similarly, the sum of the vectors b_{1}+b_{2} is a sum vector of the subspace B.

$$ \vec{a_1} + \vec{a_2} \in A $$

$$ \vec{b_1}+\vec{b_2} \in B $$

The sum of a vector from A with a vector from B belongs to the set A+B

$$ \vec{u} + \vec{w} = ( \vec{a_1} + \vec{a_2} ) + ( \vec{b_1} + \vec{b_2}) \in A+B $$

Therefore, the set A+B is closed with respect to addition.

This is the first property of vector subspaces.

**2) Scalar Multiplication Property**

Given a scalar k from K and the sum u+w of two generic elements from A+B.

$$ k \in K $$

$$ \vec{u}+\vec{w} \in A+B $$

where u and w are vectors obtained by summing a vector from A with a vector from B

$$ \vec{u} = \vec{a_1} +\vec{b_1} $$

$$ \vec{w} = \vec{a_2} + \vec{b_2} $$

with

$$ \vec{a_1}, \vec{a_2} \in A $$

$$ \vec{b_1}, \vec{b_2} \in B $$

The scalar multiplication of k with the sum of elements

$$ k ( \vec{u} + \vec{w} ) = k ( \vec{a_1} + \vec{b_2} ) + k ( \vec{a_2} + \vec{b_2} ) $$

can be rewritten as

$$ k ( \vec{u} + \vec{w} ) = k \vec{a_1} + k \vec{b_1} + k \vec{a_2} + k \vec{b_2} $$

or

$$ k ( \vec{u} + \vec{w} ) = k ( \vec{a_1} + \vec{a_2} ) + k ( \vec{b_1} + \vec{b_2} ) $$

where

$$ k (\vec{a_1}+\vec{a_2}) \in A $$

$$ k (\vec{b_1}+\vec{b_2}) \in B $$

Thus

$$ k ( \vec{u} + \vec{w} ) \in A+B $$

The scalar product ka belongs to the set A+B.

Therefore, the set A+B is closed with respect to scalar multiplication.

This also satisfies the second property of vector subspaces.

Since both properties of vector subspaces are met, it follows that **the sum of two vector subspaces is a vector subspace**.

__The sum of the subspaces A+B contains the union A∪B__

Next, I need to prove that the sum of the subspaces A+B contains the union of the subspaces A∪B

Every vector from subspace A can be written as the sum of the vector with the zero vector

$$ \forall \vec{a} \in A \ \Rightarrow \vec{a} = \vec{a} + \vec{0} $$

Since the sum A+B is a vector subspace, it contains the zero vector.

Therefore, the subspace A is contained within the sum A+B of the subspaces

$$ A \subset A+B $$

The same logic applies to every vector from the subspace B

$$ \forall \vec{b} \in B \ \Rightarrow \vec{b} = \vec{b} + \vec{0} $$

Thus, the subspace B is also contained within the sum A+B of the subspaces

$$ B \subset A+B $$

In conclusion, the sum of the subspaces A+B includes all vectors from A and B, i.e., it contains the union of the subspaces A∪B.

$$ A \cup B \subset A+B $$

## A Practical Example

Consider the vector space V=R^{3}

$$ V = R^3 $$

and two vector subspaces

$$ A = \{ \vec{a} \in R, \vec{b}=0 \} $$

$$ B = \{ \vec{a} = 0, \vec{b} \in R \} $$

The subspace A consists of vectors along the x-axis.

The subspace B consists of vectors along the y-axis.

Now, consider the subspace formed by the sum of the vector subspaces A+B

$$ A+B $$

The sum subspace A+B encompasses both subspaces A and B.

Moreover, as a subspace, it includes any vector sum and any scalar multiplication of a vector, effectively the entire plane.

## The Smallest Subspace Containing A and B

The sum of vector spaces A+B is the smallest vector subspace of V that contains the union of the subspaces A∪B.

**Proof**

Consider any vector subspace L that contains all vectors of A and B, namely the union of the two subspaces A∪B.

$$ A \cup B \subseteq L $$

Hence, every vector from subspace A is also in L

$$ \forall \ a \in A \rightarrow a \in L $$

The same holds true for vectors from B

$$ \forall \ b \in B \rightarrow b \in L $$

As L is a vector subspace, by definition, it also contains the sum of vectors from A and B

$$ \{ \vec{a}+\vec{b} \ | \ \vec{a} \in A \ , \ \vec{b} \in B \} \subseteq L $$

Thus, the vector subspace L includes the vector subspace A+B

$$ A+B \subseteq L $$

In general, any vector subspace containing A and B also includes their sum A+B.

Therefore, **among all vector subspaces that contain the union of subspaces A∪B, the sum subspace A+B is the smallest**.

**Note**. The subspace of the sum of subspaces A+B is contained in all subspaces that contain A and B. Consequently, A+B is definitely the smallest among all subspaces containing A and B. Furthermore, any subset smaller than A+B that contains A and B, $$ A, B \subset Z \subset A+B $$, is not a subspace because it does not satisfy one of the properties of vector subspaces, as the sum of vectors from A and B is not an internal operation in Z. Therefore, A+B is the smallest subspace capable of containing the subspaces A and B.

And so on