0

Given two subspaces, $$ and $$, of the same dimension within $\mathbb{R}^6$, I do not understand how to find the possible values of $\operatorname{dim}( ∩ )$.

From this post, I understand the dimension of subspaces cannot be bigger than the one containing them.

Say $\operatorname{dim}() = \operatorname{dim}() = 5$. Then how would I use the following fact? Where does the inequality come from?

$\operatorname{dim}( + ) = \operatorname{dim}() + \operatorname{dim}() − \operatorname{dim}( ∩ )$.

amWhy
  • 209,954
Hikiki Me
  • 51

2 Answers2

2

Let $d=\dim(U\cap V)$, $p=\dim U=\dim V,$ and $n=\dim E$ (the ambient space). Then, $$2p-d=\dim U+\dim V-\dim(U\cap V)=\dim(U+V)\le n,$$ i.e. $$d\ge2p-n.$$ This is the only constraint on $d$, apart of course from being $\ge0$ and $\le p.$ Hence the possible values for $d$ are: $$m,m+1,m+2,\dots,p,\quad\text{where}\quad m=\max(0,2p-n).$$ In your example: $n=6,$ $p=5,$ and $m=2p-n=4.$

Anne Bauval
  • 34,650
  • What property of subspaces tells you n = 6 (i.e., dim E)? Is it the fact that U and V are contained in R6? Are "vector space" and "ambient space" interchangeable? – Hikiki Me Oct 23 '22 at 00:40
  • 1
    @HikikiMe Yes and yes. The ambient space $E$ is the given vector space of which $U$ and $V$ are subspaces. In the example $E=\mathbb R^6.$ – Anne Bauval Oct 23 '22 at 05:25
1

If $\hbox{dim}U=\hbox{dim}V=5$, then $\hbox{dim}(U+V)$ is either $5$ or $6$, hence by the formula you quoted, $\hbox{dim}(U\cap V)$ is either $4$ or $5$. In general, if $\hbox{dim}V=\hbox{dim}U=x$, then $\hbox{dim}(U+V)$ is one of the numbers $x,x+1,\dots,\min\{2x,6\}$, and so $\hbox{dim}(U\cap V)$ can be any number between $2x-\min\{2x,6\}$ and $x$.

uniquesolution
  • 18,541