How do determine whether a subset is a subspace?

Question

So I have

\begin{align*} A&= \{(x_1,x_2)'\in \mathbb{R}^2: \max (|x_1|,|x_2|) \leq 1\} \\ B &= \{(x_1,x_2,x_3,x_4)'\in \mathbb{R}^4:x_1⋅x_2=0\} \\ C&=\{(x_1,x_2,x_3)' \in \mathbb{R}^3 :x_2-x_3-1=0 \} \end{align*}

And i want to determine whether these subsets of $\mathbb{R}^n$ are vector subspaces or not.

I know that $V \subset \mathbb{R}^n$ is a subspace when : the $0$ vector is in $V$, $V$ is closed under addition and $V$ is closed under multiplication .

I also learned that if $V$ is the set of solutions to homogenoeus linear equations , then $V$ is a subspace, and if $V$ is the span of some vectors , then $V$ is a subspace.

Since I have dyscalculia, I really do not know how to apply this information in order to check whether they are subspaces or not... They don't make sense to me unfortunately . Moroever, i don't know which of these concepts i should apply to which subset.

If anybody could help me and show me how these sets are subspaces or not , in the clearest and most simple way possible, that would be great ! Thank you !

Answer 1

$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) \in A$ but $2(1,1) =(2,2) \notin A$.

$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) \in B$ and $(1,0,0,0) \in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) \notin B$.

$C$ fails to be a subspace because $(0,0,0) \notin C$.

Answer 2

$A$ is clearly not a subspace since $(1,0) \in A $ but $(1,0) + (1,0) = (2,0) \not\in A.$

$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$

For $B$ we must check

1. $0 \in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$

2. The sum of two vectors of $B$ is in $B$ ? $\forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' \in \mathbb{R}:$

   $$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
   $(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' \neq 0$

So $B$ is not a subspace.