R^4 subspace question

Printable View

Jun 24th 2011, 03:55 PM
transgalactic

R^4 subspace question

there is U,W,V subspaces of $R^{4}$ and dimU=dimV=dimW=3
prove that $U\cap V\cap W\neq{0}$
how i tried to solve it:

dim(U+V)=dimU+dimV- $dim(U\cap V)$
because U+V is a subspace of $R^{4}$ then $dim(U+V)\leq4$
so $dim(U\cap V)\geq10$
$dim(W+(U\cap V))=dimW+dim(U\cap V)-dim(W\cap U\capV)$
because $W+(U\cap V)$ is a subspace of $R^{4}$ then
$dim(W+(U\cap V))\leq4$
by inputing the previos data i get
$dim(W\cap U\cap V)\geq17$
is it correct?
Jun 24th 2011, 08:26 PM
FernandoRevilla

Re: R^4 subspace question

Quote:

Originally Posted by transgalactic

$dim(W\cap U\cap V)\geq17$ is it correct?

Impossible, $W\cap U\cap V\subset \mathbb{R}^4$ so, $\dim (W\cap U\cap V)\leq 4$ .
Jun 24th 2011, 10:21 PM
Deveno

Re: R^4 subspace question

indeed, we have 4 = dim(R^4) ≥ dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 6 - dim(U∩V).

this means that dim(U∩V) ≥ 6-4 = 2.

now we also have 4 ≥ dim((U∩V)+W) = dim(U∩V) + dim(W) - dim((U∩V)∩W) ≥ 5 - dim(U∩V∩W).

this means that dim(U∩V∩W) ≥ 5-4 = 1.
Jun 24th 2011, 11:57 PM
transgalactic

Re: R^4 subspace question

ok i understand now
thanks

All times are GMT -8. The time now is 02:26 AM.

Copyright © 2005-2017 Math Help Forum. All rights reserved.