# Finite dimensional vector space with subspaces [duplicate]

Possible Duplicate:
Could intersection of a subspace with its complement be non empty.

Is it possible for a finite dimensional vector space to have 2 disjoint subspaces of the same dimension ? Any help would be much appreciated.

## marked as duplicate by Arturo Magidin, Nate Eldredge, Norbert, Hagen von Eitzen, davidlowryduda♦Oct 9 '12 at 17:25

• What do you mean by disjoint? Of course it's not as disjoint in the usual sense (for two sets) since a subspace should at least contain $0$. – Davide Giraudo Feb 8 '12 at 20:16
As David Giraudo points out, any subspace $U\subseteq V$ is going to contain the zero vector (it has to: the subspace is a vector space so is closed under multiplication by elements of the underlying scalar field, and zero is in the scalar field, so multiplying by it tells us the zero vector is in the subspace). In this way, no two vector subspaces are disjoint as sets. In fact, the smallest subspace is trivial, $\{0\}$.
There is a linear notion of "disjoint" here: orthogonal. For example, both a line and a plane through the origin are subspaces, and if they intersect at a right angle they are orthogonal. More generally, two vector subspaces $U$ and $W$ are orthogonal if for every $u\in U, w\in W$, the vectors $u,w$ are perpendicular. The orthogonal complement of a subspace is the maximal orthogonal subspace to it (see Wikipedia). So saying $U,W$ are orthogonal is equivalent both $U\subseteq W^\perp$ and $W\subseteq U^\perp$.