Attack the problem directly: suppose is a subspace, then:

1) If has one unique element it must be the zero vector and thus

2) If has at least one non-zero element look at ; if then is a line (through the origin, of course), otherwise

there exists is linearly independent and thus , but of course

and is a plane (again, through the origin)...in fact, then which in fact is a plane itself.

Tonio