Show that vectors [ x y z] where $x+y+z=0$ is a subspace $V$ of $\mathbb{R}^3$.

Question

Show that vectors [x y z] where $x+y+z=0$ is a subspace $V$ of $\mathbb{R}^3$.

I am having trouble understanding how to properly set this problem up. I know the properties required to be met to be considered a subspace, I just am unsure how [x y z] gets set up in this problem. It says these are vectors, plural, but there are only 3 variables so this is throwing me off. Thanks for any help!

Answer 1

If I understand correctly these are just 3D vectors where the sum of their coordinates is 0. e.g. (0,0,0), (1,-1,0),(2,-1,-1)

You can then check if the subspace properties hold for such vectors.

Answer 2

The set that we are considering here is $$W:=\{(x,y,z)\in \mathbb{R}^3\;|\;x+y+z=0\}$$ that is the set of ALL the vectors $(x,y,z)$ such that $x+y+z=0$.

In order to show that $W$ is a subspace of $\mathbb{R}^3$, you should show that it is closed with respect to linear combinations, that is if $(x_{1},y_{1},z_{1})$ and $w(x_{2},y_{2},z_{2})$ are in $W$ then also $$a(x_{1},y_{1},z_{1})+b(x_{2},y_{2},z_{2})=(ax_1+b x_2,ay_1+b y_2,az_1+b z_2)\in W$$ for all $a,b\in\mathbb{R}.$ This is equivalent to say that if $x_1+y_1+z_1=0$ and $x_2+y_2+z_2=0$ then $$(ax_1+b x_2)+(ay_1+b y_2)+(az_1+b z_2)=0$$ which can be easily verified.

Answer 3

Observe that the vectors $s:=\langle 1,0,-1\rangle$ and $t:=\langle 1,-1,0\rangle$ are linearly independent vectors in $W:= \{\textbf{x} = \langle x,y,z \rangle: x+y+z = 0\}$ i.e $W= \textbf{span}\{s,t\}\Rightarrow W$ is a vector space.

Answer 4

One way to prove that a given nonempty subset $V$ of a vector space is a subspace is to use the one-step vector subspace test. It says that $V$ is a subspace if and only if $u+\lambda v\in V$ whenever $u, v\in V$ and $\lambda\in\Bbb R$.

Note that your $V$ is a nonempty subset of $\Bbb R^3$ since $(0,0,0)\in V$. To use the one-step vector subspace test, suppose $(a,b,c),(p,q,r)\in V$ and let $\lambda\in\Bbb R$. Then $$ (a,b,c)+\lambda(p,q,r)=( \underbrace{a+\lambda p}_{=x}, \underbrace{b+\lambda q}_{=y}, \underbrace{c+\lambda r)}_{=z} $$ But this vector $(x,y,z)$ satisfies \begin{align*} x+y+z &= (a+\lambda p)+(b+\lambda q)+(c+\lambda r) \\ &= \underbrace{a+b+c}_{=0}+\lambda\cdot\underbrace{(p+q+r)}_{=0} \\ &= 0 \end{align*} That is, $(a,b,c)+\lambda(p,q,r)\in V$. Hence $V$ is a subspace.

Answer 5

Obviously $f\colon\mathbb{R}^3\to \mathbb{R}$, $f(x,y,z):=x+y+z$ is linear and $W$ is as the kernel of $f$ a subspace of $\mathbb{R}^3$.