To prove a subset is a subspace of a vector space we have to prove that the same operations (closed under vector addition and closed under scalar multiplication) on the Vector space apply to the subset. Fine, I get this.
But I am having trouble with the subspace tests. For example, if the question is:
Let $W = \{(a,b,c)|a\geq b\}$ be a subset of the vector space $V$. Show that $W$ is a subspace of $V$.
I can do upto this:
Let $w$ belong to $V$, and $w_1 = (a_1,b_1,c_1)$ and $w_2 = (a_2,b_2,c_2)$ belong to $w$. We have to show that $w_1 + w_2$ is closed under vector addition. Therefore, $w_1 + w_2 = (a_1 + a_2,b_1 + b_2,c_1 +c_2)$.
My question is, considering what the question is asking, basically how do I solve this? How to test whether it is closed under scalar multiplication when there are two vectors involved? How do I bring in $a\geq b$ in the answer?
Thank you.
Best Answer
To show a subset is a subspace, you need to show three things:
To show 1, as you said, let $w_{1} = (a_{1}, b_{1}, c_{1})$ and $w_{2} = (a_{2}, b_{2}, c_{2})$. Suppose $w_{1}$ and $w_{2}$ are in our subset. Then they must satisfy $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$. We need to check that $w_{1} + w_{2}$ is in our subset, that is, we need to check that if
$$w_{1} + w_{2} = (a_{1} + a_{2}, b_{1} + b_{2},c_{1} + c_{2})$$
then it satisfies that the first coordinate is greater than or equal to the second coordinate. Is $a_{1} + a_{2} \geq b_{1} + b_{2}$? Yes, by adding the two inequalities $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$ together. So $w_{1} + w_{2}$ is in our subset.
For 2, we need to show that if $\alpha$ is any scalar, then if $w_{1} = (a_{1}, b_{1}, c_{1})$ is in our subset, so is $\alpha w_{1}$. But $\alpha w_{1} = (\alpha a_{1}, \alpha b_{1}, \alpha c_{1})$. We know since $w_{1}$ is in our subset that $a_{1} \geq b_{1}$. We need to check that for any $\alpha$, $\alpha a_{1} \geq \alpha b_{1}$. This inequality is definitely true if $\alpha \geq 0$, but we need it to be true for all $\alpha$, and it's not, because if $\alpha$ is negative, then multiplying the inequality $a_{1} \geq b_{1}$ on both sides by a negative number makes the inequality flip. So we would get $\alpha a_{1} \leq \alpha b_{1}$ if $\alpha < 0$, which means the inequality doesn't hold. Thus, if $\alpha < 0$, then $\alpha w_{1}$ is not in our subset because it doesn't satisfy $\alpha a_{1} \geq \alpha b_{1}$.
So our subset is not a subspace because it doesn't satisfy 2 (it is not closed under scalar multiplication, because any negative scalar would cause this problem).