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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?

Con
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autodidacti
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  • If $E\cap F=\varnothing$, why does your formula involve $\dim(E\cap F)$? – Rushabh Mehta May 10 '21 at 14:42
  • In your example you have $\dim E = \dim F$ and all linear combinations of points in $E$ and $F$ are inside $F$, so we have $(E,F)=F$. Remark: It would be better to write $\mathbb{R}^3$ instead of $R^3$ – LegNaiB May 10 '21 at 14:42
  • I managed to show that $(E,F)=F$ algebraically, but then it seems the second conclusion of the theorem does not hold? – autodidacti May 10 '21 at 15:04

1 Answers1

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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?
lonza leggiera
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  • First, I made a mistake when reaching the conclusion that $(E,F)=F$, so thank you for helping me realize that. Second, since a,b, and x are arbitrary real numbers, I guess the set of points of the form $(a,b,2x)$ is exactly $\Bbb{R^3}$, which confirms both cases. Is this correct? – autodidacti May 10 '21 at 15:45
  • Yes, you've got it. – lonza leggiera May 10 '21 at 17:47