Subspace Addition Proof

Ah right, so is L+M = R^2?

Am I right in saying L+N is also R^2? So the RHS is R^2 with L on the LHS giving a counter-example?

Correct.

I don't know the definition you've been given, but there may have been slightly easier ways to get L+N=R². I'm going to take the definition L+N=span(L_UN).

Intuitive: You have two non-parallel lines through the origin. You can clearly reach quite a few points by traveling along L followed by a line parallel to N. Thinking for a bit longer you can see that you can actually reach any point this way.

Formal: Unfortunately I've forgotten all the results. Maybe something like if X is a set of linearly independent vectors then X is a basis for span(X), and by definition the dimension is the cardinality of the basis. Then something about the dimension of a proper subspace of a finite dimensional vector space. There could well be something more direct.

(edited 11 years ago)

Reply 21

ghostwalker

17

Original post by Noble.

Ah right, so is L+M = R^2?

Am I right in saying L+N is also R^2? So the RHS is R^2 with L on the LHS giving a counter-example?

Yep - as harr said.

Reply 22

Noble.

OP

17

Original post by harr

Correct.

I don't know the definition you've been given, but there may have been slightly easier ways to get L+N=R². I'm going to take the definition L+N=span(L_UN).

Intuitive: You have two non-parallel lines through the origin. You can clearly reach quite a few points by traveling along L followed by a line parallel to N. Thinking for a bit longer you can see that you can actually reach any point this way.

Formal: Unfortunately I've forgotten all the results. Maybe something like if X is a set of linearly independent vectors then X is a basis for span(X), and by definition the dimension is the cardinality of the basis. Then something about the dimension of a proper subspace of a finite dimensional vector space. There could well be something more direct.