Linear Algebra: A Basis for a Finite Dim VS

[Bachelier]

Why is it enough to prove that a set of vectors is a BASIS to a FINITE DIMENSIONAL Vector Space, it is enough to show that it is Linearly Independent.

No Need to prove that it spans the whole vector space?

 

[Fredrik]

Hint: What is the definition of finite-dimensional? A vector space is n-dimensional, if...

 

[Bachelier]

Hint: What is the definition of finite-dimensional? A vector space is n-dimensional, if...

...if it has a basis consisting of a finite number of vectors. In this case n vectors.
This is my point, we still have to show that the number of elements in the set is equal to the dimension of the vector space V. For instance if the dim is unknown, then we must show the set spans. Otherwise using just the fact that V is finite dimensional is not enough.
Am I reading this correctly?

Thank you.

 

[arkajad]

A basis is a set of linearly independent vectors that span all the space. The first important theorem tells us that if one basis is finite then any other basis is also finite and they have the same number of vectors. This common number is then called the dimension of the space.

 

[wisvuze]

a basis is any set of linearly independent vectors that span a space V.
If you are given a dimension for a vector space V, then by the replacement theorem , any linearly independent subset of V that's of size n will form a basis for V

 

[Fredrik]

...if it has a basis consisting of a finite number of vectors. In this case n vectors.

OK, with that definition things get circular. The one I had in mind is that dim V=n if there exists a linearly independent subset of V with n members but no linearly independent subset of V with n+1 members. Suppose that dim V=n, and that B is a linearly independent subset of V with n members. Let x be an arbitrary member of V. If x is in B, then x is obviously a linear combination of the members of B. If x is in V-B, then [itex]B\cup\{x\}[/itex] is linearly dependent (since it has n+1 members), so x is a linear combination of the members of B in this case too.

 

[Landau]

Perhaps I am reading the opening post different than you guys, but my answer: it is NOT enough, since otherwise any linearly independent set - whether or not it is maximal - would be a basis.

The set

[tex]\{(1,0,0),(0,1,0)\}\subset\mathbb{R}^3[/tex]

is linearly independent, but not a basis.

The set

[tex]\{(1,0,0),(0,1,0),(0,0,1)\}\subset\mathbb{R}^3[/tex]

is also linearly independent, but moreover it spans R^3, hence it is a basis.

Of course, if you know the dimension of the vector space, say n, then in order to prove that some set consisting of precisely n vectors is a basis, it IS enough only to show linear independence. But that was not included in the question.

 

[Fredrik]

The OP should have said "a set of n vectors" and "an n-dimensional vector space" instead of "a set of vectors" and "a finite dimensional vector space". I sort of assumed that that's what he meant, and then quickly forgot that he actually said something else.

 

[Anonymous217]

I thought the question was pretty straight-forward by itself and needs no clarification.. And yes, I agree with Landau and his counterexample. It simply isn't enough.

 

[Bachelier]

Great answers everbody. Thank you very much.