# Which of the following are subspaces of the given real vector spaces?

$$B = \{(a, b, c) \in R^3:b \geq0\} \subseteq R^3$$

$$C = \{(a - b, a+ b, 7a): a, b \in R\} \subseteq R^3$$

I know the steps to prove these are:

Let V be a vector space. A is a subset $$W \subseteq V$$ is a subspace of V if

1. W is non-empty / must contain a zero vector
2. W is closed under vector addition
3. W is closed under scalar multiplication .



Solving the first one:

$$u , v \in R^3$$ $$u, v \in B$$ $$u = (u_1, u_2, u_3),v = (v_1, v_2, v_3)$$ $$u_2, v_2 \geq 0$$ $$u_1 + v_1, u_1 + v_1,u_1 + v_1 \in R$$ $$\therefore (u_1 + v_1, u_1 + v_1, u_1 + v_1) \in R^3$$ $$(u_1 + v_1, u_1 + v_1, u_1 + v_1) = u + v$$ $$\therefore u + v \in B$$ Vector addition is closed   $$\alpha u = (\alpha u_1,\alpha u_2,\alpha u_3), u_2 \geq0$$ $$\alpha,u_1, u_2, u_3 \in R$$ $$\therefore \alpha u_1,\alpha u_2,\alpha u_3 \in R$$ $$\therefore (\alpha u_1,\alpha u_2,\alpha u_3) \in R$$ and so $$\alpha u \in B$$ and Scalar multiplication is closed

 I hope these are correct, but how do I do C? From observation, shouldn't any vector $$(a,b,c)$$ where $$a,b,c \in R$$ always be in $$R^3$$, how come I need to do this?

• For scalar multiplication on $B$, what if $a<0?$ Commented Sep 8, 2019 at 12:41
• $C$ is contained in $\Bbb R^3,$ so sums of elements of $C$ are in $\Bbb R^3,$ but you have to show they’re in $C$ Commented Sep 8, 2019 at 12:44
• @J.W.Tanner This means that $u_2 \le 0$! and so is not closed under scalar multiplication, so its not a subspace Commented Sep 8, 2019 at 12:48
• that’s correct! Commented Sep 8, 2019 at 12:51

With $$\;B\;$$ check what happens with $$\;\alpha=-1\;$$ ...
For $$\;C\;$$ : do exactly as you did with $$\;B\;$$ . For example, addition:
$$(a-b,\,a+b,\,7a)+(\alpha-\beta,\,\alpha+\beta,\,7\alpha)=\left((a+\alpha)-(b+\beta),\,(a+\alpha)+(b+\beta), 7(a+\alpha)\right)$$