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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.

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    $\begingroup$ This isn't closed under scalar multiplication $(1,-1,0)\in W$ yet $-1\cdot(1,-1,0) = (-1,1,0)$ is not. $\endgroup$
    – James
    Jul 18, 2014 at 18:48

2 Answers 2

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To show a subset is a subspace, you need to show three things:

  1. Show it is closed under addition.
  2. Show it is closed under scalar multiplication.
  3. Show that the vector $0$ is in the subset.

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).

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  • $\begingroup$ Thanks, but when you say "Yes, by adding the two inequalities a1≥b1 and a2≥b2 together. " where do you get the inequalities from. $\endgroup$
    – studious
    Jul 18, 2014 at 20:25
  • $\begingroup$ Also, sorry but I am having difficulty applying the same ideas to this question, Show that W = {(a,b,c)|a^2+b^2+c^2 <= 1} is not a subspace of V, where W is a subset of V. How to do take w1= (a1,b1,c1) to a1^2+b1^2+c1^2 <= 1? $\endgroup$
    – studious
    Jul 18, 2014 at 20:28
  • $\begingroup$ @pilot We know $w_{1} = (a_{1},b_{1},c_{1})$ is in our subset, so this gives us $a_{1} \geq b_{1}$. Also, $w_{2} = (a_{2},b_{2},c_{2})$ is in our subset, and this gives us $a_{2} \geq b_{2}$. So we know that $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$ because $w_{1}$ and $w_{2}$ are in the subset. Adding the two inequalities gives us $a_{1} + a_{2} \geq b_{1} + b_{2}$, which means the vector $w_{1} + w_{2}$ is in our subset, since it satisfies the property that its first coordinate is greater than or equal to its second coordinate. $\endgroup$
    – layman
    Jul 18, 2014 at 21:20
  • $\begingroup$ @pilot To show $W = \{ (a, b, c) \mid a^{2} + b^{2} + c^{2} \leq 1 \}$ is not a subspace of $V$, we need to show that at least one of the three subspace conditions does not hold. Clearly, $(0, 0, 0)$ is in $W$ since $0^{2} + 0^{2} + 0^{2} \leq 1$. So condition 3 holds. So either condition 1 or 2 does not hold. I think condition 2 is easier to check, so we will check it first. Is it true that for all scalars $\alpha$, if $u = (a, b, c)$ is in $W$, then $\alpha u$ is in $W$? We need to check that for all $\alpha$...(continued below) $\endgroup$
    – layman
    Jul 18, 2014 at 21:24
  • $\begingroup$ @pilot we need to check that for all scalars $\alpha$, if we know $a^{2} + b^{2} + c^{2} \leq 1$, is it true that $(\alpha a)^{2} + (\alpha b)^{2} + (\alpha c)^{2} \leq 1$ (for all $\alpha$)? The answer is no. By factoring out an $\alpha^{2}$ from the inequality and dividing both sides by $a^{2} + b^{2} + c^{2}$, we see that the only $\alpha$ that satisfy this equation are those such that $\alpha^{2} \leq \dfrac{1}{a^{2} + b^{2} + c^{2}}$. So picking an $\alpha$ such that $\alpha^{2} > \dfrac{1}{a^{2} + b^{2} + c^{2}}$ means $\alpha u$ does not satisfy the condition, so it is not in $W$. $\endgroup$
    – layman
    Jul 18, 2014 at 21:27
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Do you suspect this is a vector space or not? If you think it is, you have to verify the definitions. If not, then find counterexamples.

Testing for scalar multiplication doesn't require two vectors, you examine the result of multiplying one vector by one scalar.

Hint:

Try multiplying by the scalar $-1$, let $v = -w$ with $w \in W$. Is $v$ in $W$ also (i.e. is $-a \geq -b$?)

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