# Vector Sum Theorems

We will now look at two very important theorems regarding the sum of vector subspaces. The first theorem will tell us that the sum of any subspaces of $V$ will result in a subspace of $V$, and this sum will contain all of the subspaces in its sum. The second theorem will tell us that $(U_1 + U_2)$ is the smallest subspace containing both $U_1$ and $U_2$.

Theorem 1: Let $V$ be a vector space over the field $\mathbb{F}$ and let $U_1, U_2$ be subspaces of $V$. Then $(U_1 + U_2)$ is a subspace of $V$, and $(U_1 + U_2)$ contains $U_1$ and $U_2$. |

**Proof:**We first note that $(U_1 + U_2) \subseteq V$ since all vectors in $U_1$ are in $V$ and all vectors in $U_2$ are in $V$. We want to show that $(U_1 + U_2)$ is a subspace of $V$ by verifying that $0 \in (U_1 + U_2)$, and that $(U_1 + U_2)$ is closed under addition and scalar multiplication.

- First, we have that $0 \in U_1$ and $0 \in U_2$ since all vector subspaces must contain the zero vector. Therefore $0 = \underbrace{0}_{0 \in U_1} + \underbrace{0}_{0 \in U_2}$ and so since $0$ can be written as a sum of a vector from $U_1$ and a vector from $U_2$, it follows that $0 \in (U_1 + U_2)$.

- Let $x, y \in (U_1 + U_2)$. Then it follows that there exists vectors $x_1, y_1 \in U_1$ and $x_2, y_2 \in U_2$ such that $x = \underbrace{x_1}_{x_1 \in U_1} + \underbrace{x_2}_{x_2 \in U_2}$ and $y = \underbrace{y_1}_{y_1 \in U_1} + \underbrace{y_2}_{y_2 \in U_2}$. Therefore $x + y = (x_1 + x_2) + (y_1 + y_2) = \underbrace{(x_1 + y_1)}_{(x_1 + y_1) \in U_1} + \underbrace{(x_2 + y_2)}_{(x_2 + y_2) \in U_2}$, and so therefore $x + y$ can be written as the sum of one vector in $U_1$ and one vector in $U_2$ and so $(x + y) \in (U_1 + U_2)$, or in other words, $(U_1 + U_2)$ is closed under addition.

- Let $x \in (U_1 + U_2)$. Then there exists a vector $x_1 \in U_1$ and a vector $x_2 \in U_2$ such that $x = \underbrace{x_1}_{x_1 \in U_1} + \underbrace{x_2}_{x_2 \in U_2}$. Let $a \in \mathbb{F}$. If we multiply both sides by $a$ we get that $ax = a(x_1 + x_2) = ax_1 + ax_2$. But $ax_1 \in U_1$ since $U_1$ is a subspace and closed under scalar multiplication. Similarly $ax_2 \in U_2$ since $U_2$ is a subspace and closed under scalar multiplication. So $ax = \underbrace{ax_1}_{ax_1 \in U_1} + \underbrace{ax_2}_{ax_2 \in U_2}$ can be written as the sum of a vector from $U_1$ and a vector from $U_2$ and so $ax \in (U_1 + U_2)$ or in other words, $(U_1 + U_2)$ is closed under scalar multiplication.

- Therefore $(U_1 + U_2)$ is a vector subspace.

- Lastly we need to show that $(U_1 + U_2)$ contains both $U_1$ and $U_2$, that is $U_1 \subseteq (U_1 + U_2)$ and $U_2 \subseteq (U_1 + U_2)$. Let $x \in U_1$. We have that $0 \in U_2$ by the definition of a vector space and so $x = \underbrace{x}_{x \in U_1} + \underbrace{0}_{0 \in U_2}$ and so $x$ can be written as the sum of a vector from $U_1$ and $U_2$ and so $x \in (U_1 + U_2)$. Therefore $U_1 \subseteq (U_1 + U_2)$. Similarly let $y \in U_2$. We have that $0 \in U_1$ by the definition of a vector space and so $y = \underbrace{0}_{0 \in U_1} + \underbrace{y}_{y \in U_2}$ and so $y$ can be written as the sum of a vector from $U_1$ and $U_2$ and so $y \in (U_1 + U_2)$. Therefore $U_2 \subseteq (U_1 + U_2)$. $\blacksquare$

Theorem 2: Let $V$ be a vector space over the field $\mathbb{F}$ and let $U_1$ and $U_2$ be subspaces of $V$. If $U$ is a subspace of $V$ such that $U_1 \subseteq U$ and $U_2 \subseteq U$ then $(U_1 + U_2) \subseteq U$. |

**Proof:**Let $U$ be a subspace of $V$ such that $U_1 \subseteq U$ and $U_2 \subseteq U$. We want to show that $(U_1 + U_2) \subseteq U$.

- First let $x \in (U_1 + U_2)$. Therefore there exists an $x_1 \in U_1$ and an $x_2 \in U_2$ such that $x = \underbrace{x_1}_{x_1 \in U_1} + \underbrace{x_2}_{x_2 \in U_2}$.

- Now since $U_1 \subseteq U$ we have that since $x_1 \in U_1$ then this implies that $x_1 \in U$. Similarly since $U_2 \subseteq U$ we have that since $x_2 \in U_2$ then this implies that $x_2 \in U$. Now since $U$ is a subspace of $V$, we have that $U$ is closed under addition and so $x_1 + x_2 = x \in U$.

- Therefore since $x \in (U_1 + U_2)$ implies that $x \in U$, we have that $(U_1 + U_2) \subseteq U$. $\blacksquare$

## Example 1

**Suppose that $U_1$, $U_2$, and $U_3$ are subspaces of the vector space $V$ over the field $\mathbb{F}$, and suppose that $U_1 + U_2 = U_1 + U_3$. Is it true that then $U_2 = U_3$?**

To understand this problem, we need to have a clear understanding of what a subspace sum is. We note that $(U_1 + U_2) := \{ u_1 + u_2 : u_1 \in U_1 \: \mathrm{and} \: u_2 \in U_2 \}$ and $(U_1 + U_3) := \{ u_1 + u_3 : u_1 \in U_1 \: \mathrm{and} \: u_3 \in U_3 \}$.

In other words, if $(U_1 + U_2) = (U_1 + U_3)$, then must it be that any $x \in (U_1 + U_2) = (U_1 + U_3)$, that $x = u_1 + u_2 = u_1 + u_3$, in other words, $u_2 = u_3$?

The answer is no. Suppose that $V$ is a nonzero vector space, and let $U_1 = V$, let $U_2 = V$, and let $U_3 = \{ 0 \}$. Then we have that:

(1)

Clearly $V + V = V + \{ 0 \}$ since any vector $x \in V$ can be written as $x = \underbrace{x_1}_{\in V} + \underbrace{x_2}_{\in V}$ and any vector $x \in V$ can be written as $x = \underbrace{x_1}_{\in V} + \underbrace{\in 0}$, but $V \neq \{ 0 \}$ since we asserted $V$ was a nonzero vector space.