Dimension and Basis of a Vector Subspace

Question

Given $F=\{(x,y,z,t)\in \mathbb{R}^4: x+t=0\}$ and $G=\{(x,y,z,t)\in \mathbb{R}^4:y=-z, x=t\}$, how can you prove they are subspaces of $\mathbb{R}^4$? Besides, if you want to find the basis and the dimension of $F$, $G$, $F\cap G$ and $F\cup G$, how would you proceed?

Answer 1

It is helpful to rewrite the elements of $F$ and $G$ so that their elements are clearer. For example, $$F = \{(x,y,z,-x)\;|\;x,y,z \in \mathbb{R}\}\;\text{and}\;G = \{(x,y,-y,x)\;|\;x,y \in \mathbb{R}\}.$$ I'll leave it to you to show that $F$ and $G$ are subspaces of $\mathbb{R}^4$ and that $\{(1,0,0,-1), (0,1,0,0), (0,0,1,0)\}$ is a basis of $F$ and that $\{(1,0,0,1), (0,1,-1,0)\}$ is a basis of $G$. The thought process behind finding these is the idea that $F$ depends only on the first three variables whereas $G$ depends only on the first two by definition.

We need to be careful about $F \cup G$ because it is not actually a subspace. For example, $(1,0,0,1), (1,0,0,-1) \in F \cup G$ whereas their sum $(2,0,0,0)$ is not. $F \cap G$ is a subspace and it's given by $$F \cap G = \{(0,y,-y,0)\;|\;y \in \mathbb{R}\}$$ with a one element basis $\{(0,1,-1,0)\}$.