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

linear algebravector-spacesvectors

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.

Best Answer

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. :)

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