# Thread: Proving intersection is a subspace

1. ## Proving intersection is a subspace

Let $\displaystyle W_1$ and $\displaystyle W_2$ be two subspaces of a vector space $\displaystyle V$ over the field $\displaystyle \mathbb{F}$. Prove that the intersection of $\displaystyle W_1$ and $\displaystyle W_2$ is a subspace of $\displaystyle V$.

Can someone check my working for this question

Let $\displaystyle x\in W_1, y\in W_1$

Since $\displaystyle W_1$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_1$

Similarly, let $\displaystyle x\in W_2, y\in W_2$

Since $\displaystyle W_2$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_2$

$\displaystyle \Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition

Let $\displaystyle \lambda\in\mathbb{R}, x\in W_1, W_2$

$\displaystyle \lambda x\in W_1$ since $\displaystyle W_1$ is a subspace

$\displaystyle \lambda x\in W_2$ since $\displaystyle W_2$ is a subspace

Clearly, $\displaystyle \lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication

$\displaystyle \Therefore W_1 \cap W_2$ is a subspace of $\displaystyle V$

What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?

2. Originally Posted by acevipa
Let $\displaystyle W_1$ and $\displaystyle W_2$ be two subspaces of a vector space $\displaystyle V$ over the field $\displaystyle \mathbb{F}$. Prove that the intersection of $\displaystyle W_1$ and $\displaystyle W_2$ is a subspace of $\displaystyle V$.

Can someone check my working for this question

Let $\displaystyle x\in W_1, y\in W_1$

Since $\displaystyle W_1$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_1$

Similarly, let $\displaystyle x\in W_2, y\in W_2$

Since $\displaystyle W_2$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_2$

$\displaystyle \Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition

Let $\displaystyle \lambda\in\mathbb{R}, x\in W_1, W_2$

$\displaystyle \lambda x\in W_1$ since $\displaystyle W_1$ is a subspace

$\displaystyle \lambda x\in W_2$ since $\displaystyle W_2$ is a subspace

Clearly, $\displaystyle \lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication

$\displaystyle \Therefore W_1 \cap W_2$ is a subspace of $\displaystyle V$

What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?
well, since $\displaystyle \displaystyle W_1$ and $\displaystyle \displaystyle W_2$ are subspaces, they each contain 0. so 0 will be in their intersection, and hence, the intersection is non-empty.

containing zero is important for vector spaces and are hence important for subspaces. so the subspaces must have 0 in them. it is not to say that they have to have 0 to be non-empty per sae, but it is usually easy to verify that the zero vector is in there. so you always check that. and if 0 is not in there, then you're in trouble. you wouldn't have a vector space at all. just some other kind of set.

3. Thanks a lot. I understand that well now.