I have this question from a previous problem:
Hyperplanes in R5 have dimension 4. For 3 such hyperplanes in general position, what is the dimension of their intersection?
To me, it seems the dimension of their intersection is 3, however, it's wrong. Does anyone has any hint?
Many thanks
Given distinct $\Bbb R^5$ hyperplanes $U,V,W$ then the dimension $d:=\dim(U\cap V\cap W)$ of their intersection can be $2$ or $3$ depending on $U,V,W$.
$$5=\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V)=8-\dim(U\cap V)$$ $$\therefore \dim(U\cap V)=3\;\;\;\;\therefore U\cap V\subseteq W\implies d=\dim(U\cap V)=3$$ If $U\cap V$ is not wholly contained in $W$ then $\Bbb R^5=(U\cap V)+W$ and we'd have $$5=\dim((U\cap V)+W)=\dim(U\cap V)+\dim(W)-d=7-d$$ $$\therefore\;\; d=2\;\lor\;d=3$$