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

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

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?

  • 1
    $\begingroup$ For scalar multiplication on $B$, what if $a<0?$ $\endgroup$ Commented Sep 8, 2019 at 12:41
  • 1
    $\begingroup$ $C$ 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 $C$ $\endgroup$ Commented Sep 8, 2019 at 12:44
  • $\begingroup$ @J.W.Tanner This means that $u_2 \le 0$! and so is not closed under scalar multiplication, so its not a subspace $\endgroup$ Commented Sep 8, 2019 at 12:48
  • $\begingroup$ that’s correct! $\endgroup$ Commented Sep 8, 2019 at 12:51

1 Answer 1



With $\;B\;$ check what happens with $\;\alpha=-1\;$ ...

For $\;C\;$ : do exactly as you did with $\;B\;$ . For example, addition:

$$(a-b,\,a+b,\,7a)+(\alpha-\beta,\,\alpha+\beta,\,7\alpha)=\left((a+\alpha)-(b+\beta),\,(a+\alpha)+(b+\beta), 7(a+\alpha)\right)$$


  • $\begingroup$ Just to check, I had made some edits to the logic for B in the vector addition section (lines 5 & 6) of the vector addition part. This is a suitable statement? @DonAntonio $\endgroup$ Commented Sep 8, 2019 at 13:01
  • 1
    $\begingroup$ @PoChenLiu For vector addition is fine, yet for scalar multiplication is still wrong...Read my answer $\endgroup$
    – DonAntonio
    Commented Sep 8, 2019 at 13:16

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