Dimension of the smallest affine subspace that contains two distinct affine subspaces

Question

Notation : (E,F) denotes the smallest affine subspace of $R^n$ that contains the affine subspaces $E$ and $F$.

Theorem : Let $E$ and $F$ be affine subspaces of $R^n$. Assume that $E\cap{F}=\emptyset$. Then

1. $dim(E,F)<dim(E)+dim(F)-dim(E\cap{F})$
2. $max(dim(E),dim(F)) + 1 \le{dim(E,F)}$

I want to construct an example for both cases in the preceding theorem in $R^3$. I considered the $xy$-plane and the plane $z=2$ in $R^3$. That is, I considered the affine subspace (also a vector subspace) E={$(a,b,0) : a,b \in{R}$} and the affine subspace $F=(0,0,2) + E$. But I get stuck when it comes to the dimension of $(E,F)$, I cannot imagine all the linear combinations of points in $E$ and $F$ geometrically and cannot define this set algebraically. Any ideas? Also, how should I define the dimension of the empty set?

Answer 1

Hints

• With $\ E\ $ and $\ F\ $ as you've defined them it's not true that $\ (E,F)=F\ $.
• If $\ \alpha, \beta\ $ are any points in an affine subspace, $\ A\ $ then all points on the line $\ \{\,x\alpha+(1-x)\beta\,|\,x\in\mathbb{R}\,\}\ $ must lie in $\ A\ $. The affine space $\ (E,F)\ $ contains all points of the form $\ (a,b,0)\ $, because these are all in $\ E\ $, and all points of the form $\ (a,b,2)\ $, because these are all in $\ F\ $. Therefore it must contain all points of the form $ x(a,b,2)+$$(1-x)(a,b,0)=(a,b,2x)\ $. What space does that give you?