The Intersection and Union of Subspaces

Table of Contents

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

U_1 \supseteq U_2

$\Rightarrow$ 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 $u_1 \in U_1 \setminus U_2$ .

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