Question

Show that, in the vector space V = R2, the subset of all vectors whose entries sum to zero is a subspace, but the subset of all vectors whose entries sum to one is not a subspace.

Answer 1

Question 1.

Show that, in the vector space $V=\mathbb{R}^{2}$, the subset of all vectors whose entries sum to zero is a subspace, but the subset of all vectors whose entries sum to one is not a subspace.

Solution. Denote

$A=\{v=(x,y)\in V\mid x+y=0\}$

and

$B=\{v=(x,y)\in V\mid x+y=1\}.$

Prove that $A$ is a subspace of $V$, while $B$ is not a subspace.

Indeed, for any two vectors $v_{1}=(x_{1},y_{1})$ and $v_{2}=(x_{2},y_{2})$ from $A$ and for any scalars $\alpha_{1},\alpha_{2}\in\mathbb{R}$ consider the linear combination

$\alpha_{1}v_{1}+\alpha_{2}v_{2}=(\alpha_{1}x_{1}+\alpha_{2}x_{2},\alpha_{1}y_{1}+\alpha_{2}y_{2}).$

Since

$(\alpha_{1}x_{1}+\alpha_{2}x_{2})+(\alpha_{1}y_{1}+\alpha_{2}y_{2})=\alpha_{1}(x_{1}+y_{1})+\alpha_{2}(x_{2}+y_{2})=\alpha_{1}\cdot 0+\alpha_{2}\cdot 0=0,$

we conclude that $\alpha_{1}v_{1}+\alpha_{2}v_{2}\in A$. This shows that $A$ is a subspace of $V$.

The fact that $B$ is not a subspace is obvious, because the zero vector $(0,0)$ does not belong to $B$: the sum of its coordinates is $0\neq 1$. ∎