The Intersection and Union of Subspaces

# The Intersection and Union of Subspaces

We will now look at a couple of theorems regarding the intersection of subspaces and the union of subspaces.

Theorem 1: Let $V$ be a vector space over the field $\mathbb{F}$ and let $U_1, U_2, ..., U_m$ be any collection of subspaces of $V$. Let $I$ be a set of indices denote subspaces. Then $\bigcap_{i \in I} U_i$ is a subspace. |

**Proof:**Consider the collection of subspaces $\{ U_i : i \in I$ where $I = \{ 1, 2, ..., m \}$, that is the set $\{ U_i : i \in I \}$ is a set any set of subspaces of $V$.

- Now let $x, y \in \bigcap_{i \in I} U_i$, and so $x, y \in U_i$ for all $i \in I$. Since $U_i$ is a subspace for all $i \in I$, then we know that for all scalars $a, b \in \mathbb{F}$, $(ax + by) \in U_i$ for all $i \in I$ and so $(ax + by) \in \bigcap_{i \in I} U_i$.

- Therefore the intersection of the collection of subspaces $\{ U_i : i \in I \}$ is a subspace. $\blacksquare$

Theorem 2: Let $V$ be a vector space over the field $\mathbb{F}$ and let $U_1, U_2$ be any subspaces of $V$. Then $U_1 \bigcup U_2$ is a subspace of $V$ if and only if $U_1 \subseteq U_2$ or $U_1 \supseteq U_2$. |

**Proof:**Let $U_1$ and $U_2$ be any subspaces to the vector set $V$.

- $\Rightarrow$ We first want to show that if $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$ then $U_1 \cup U_2$ is a subspace of $V$. Without loss of generality suppose that $U_1 \subseteq U_2$. Then it follows that $U_1 \cup U_2 = U_2$. But $U_2$ is defined to be a subspace of $V$ and so $U_1 \cup U_2$ is a subspace of $V$.

- $\Leftarrow$ We now want to show that if $U_1 \cup U_2$ is a subspace of $V$ then either $U_1 \subseteq U_2$ or $U_1 \supseteq U_2$. Suppose that $U_1 \not \subseteq U_2$ and $U_1 \not \supseteq U_2$. We will derive a contradiction with this.

- Let $U_1 \cup U_2$ be a subspace of $V$. Thus there exists elements $u_1 \in U_1 \setminus U_2$ and $u_2 \in U_2 \setminus U_1$. We have that $u_1, u_2 \in U_1 \cup U_2$, and since $U_1 \cup U_2$ is a subspace, then the vector $x = u_1 + u_2 \in U_1 \cup U_2$. Therefore $x \in U_1$ or $x \in U_2$.

- First consider the case where $x \in U_1$. Since $x = u_1 + u_2$ then $u_2 = x - u_1$ and so $u_2 \in U_1$ which is a contradiction since $u_2 \in U_2 \setminus U_1$.

- Now consider the case where $x \in U_2$. Since $x = u_1 + u_2$ then $u_1 = x - u_2$ and so $u_1 \in U_2$ which is a contradiction since $u_1 \in U_1 \setminus U_2$.

- Therefore either $U_1 \subseteq U_2$ or $U_1 \supseteq U_2$. $\blacksquare$