How to determine whether a set is a subspace of a particular Vector Space

Question

Given we have a set $W=\{ (x,y,z) \in \mathbb{R}^3 \;:\; x + 2y - z = 0\}$, how would I be able to determine whether it's a subspace of $\mathbb{R}^3$ ? I understand a subspace is contained within a vector space, and is closed under multiplication and addition. Can someone explain the geometrical interpretation/ relevance of the

'$x + 2y - z = 0$' part too. Thanks in advance.

Answer 1

Let $W = \{ (x,y,z) \in \mathbb{R}^3 \;:\; x+2y-z=0\}$.

So geometrically your $W$ is the set of all point in $\mathbb{R}^3$ which lie on the plane whose equation is $x+2y-z=0$. This plane passes through the origin: $0+2(0)-0=0$ (i.e., $(0,0,0) \in W$) and is perpendicular to $[1\;2\;-1]$.

To show $W$ is a subspace of $\mathbb{R}^3$ you simply need to verify that $W$ is a subset of $\mathbb{R}^3$ (this is obvious). Then you need to make sure it is non-empty (this is accomplished by noting that $(0,0,0) \in W$). Then the real task is to show it is closed under vector addition and scalar multiplication.

Take $(x,y,z), (a,b,c) \in W$ and $s \in \mathbb{R}^3$. Since $(x,y,z) \in W$, you have $x+2y-z=0$ similarly for $(a,b,c)$. Notice that adding the equation involving $x,y,z$ and the equation involving $a,b,c$ yields an equation for the vector $(x+a,y+b,z+c)$. Thus $(x,y,z)+(a,b,c)=(x+a,y+b,z+c) \in W$. Similarly, scaling the equation involving $x,y,z$ by $s$ yields an equation for $(sx,sy,sz)$. Thus $s(x,y,z)=(sx,sy,sz) \in W$.

This will show that $W$ is a subspace of $\mathbb{R}^3$.

An easier way to see that $W$ is a subspace is to note that it is the nullspace (or kernel) of a matrix/transformation whose matrix is $[1\;2\;-1]$. Notice that $$\begin{bmatrix} 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} =0$$ is exactly the criterion for belonging to $W$.

In the end, every subspace can be recognized to be a nullspace of something (or the column space/span of something).

Geometrically, subspaces of $\mathbb{R}^3$ can be organized by dimension:

• Dimension 0: The only 0-dimensional subspace is $\{(0,0,0)\}$
• Dimension 1: The 1-dimensional subspaces are lines through the origin. They are the span of a single non-zero vector. Or you can view these as the set of solutions of a pair of homogeneous equations (equations set equal to $0$).
• Dimension 2: The 2-dimensional subspaces are planes through the origin. They are the span of a pair of (linearly independent) vectors. Or you can view these are the set of solutions of a single homogeneous equation (your $W$ is exactly this).
• Dimension 3: The only 3-dimensional subspace of $\mathbb{R}^3$ is $\mathbb{R}^3$ itself.

Basically subspaces of $\mathbb{R}^n$ are "flat" things through the origin. Since your $W$ is one of those kinds of things, it's a subspace. :)

Answer 2

The set is interpreted as the set of points in $\Bbb{R}^3$, taking the form $(x, y, z)$, such that $x + 2y - z = 0$. So, for example $(\color{red}{-1}, \color{blue}1, \color{green}1)$ belongs to this set, as $\color{red}{-1} + 2 \cdot \color{blue}1 - \color{green}1 = 0$. On the other hand, the point $(\color{red}0, \color{blue}1, \color{green}1)$ does not belong to the set, as $\color{red}{0} + 2 \cdot \color{blue}1 - \color{green}1 \neq 0$.

Geometrically, we can view this equation as a dot product: $$(1, 2, -1) \cdot (x, y, z) = 0,$$ meaning that the set is the set of all points that are orthogonal to $(1, 2, -1)$. This will form a plane, passing through the origin.

How do you go about showing it's a subspace? Start by taking two vectors in the set, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Both belong to the set, so you may assume \begin{align*} x_1 + 2y_1 - z_1 &= 0 \\ x_2 + 2y_2 - z_2 &= 0. \end{align*} You need to prove that $(x_1, y_1, z_1) + (x_2, y_2, z_2)$ belongs to this set. Simplifying, we get $$(x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1 + x_2, y_1 + y_2, z_1 + z_2),$$ by definition of $+$ in $\Bbb{R}^3$. So, in order to show that this is a member of the given set, you must prove $$(x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) = 0,$$ given the two assumptions above. There are no tricks to it; the proof of closure under $+$ should only be a couple of steps away.

Then, do the same with scalar multiplication. Take $(x_1, y_1, z_1)$ as above, and let $\lambda \in \Bbb{R}$. What does it mean for $\lambda(x_1, y_1, z_1)$ to belong to the set? Prove it!

The last requirement (which some people forget to mention) is that you also have to verify the set is non-empty. Fortunately, I did that for you!