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Apologies for the grandiose title, but it is motivated by a serious consideration. Linear algebra, LA hereafter, is an enormously interesting area of mathematics. What's more, it is fairly straightforward to develop some persistently useful results; either using matrices, or in a more pure mathematical perspective, by manipulations of linear transformations.

So, I am currently reading Frames for Undergraduates by Han, Larson, et al. It starts with a review of the LA needed for the rest of the book (though clearly some of the more basic results are quickly presented and hustled past.)

One of the theorems that is presented is the following:

Theorem: Let $V$ be a finte-dimensional vector space, and suppose $\{x_{i=1}^n\}$ is a basis for $V$. Suppose $z$ is any vector in $V$ which is not a linear combination of the vectors $\{x_1,\, x_2,\, \ldots,\, x_{n-1}\}$. Then $\{x_1,\, x_2,\, \ldots,\, x_{n-1},\, z\}$ is also a basis for $V$.

Now a few things are immediate to me. One is that we have the concepts of a basis, span and linear independence, so a theorem like this is fairly straightforward.

"Proof": Given any set, $S$, of linear independent vectors, we automatically have a basis, since their span generates some subspace, $U$, of $V$. So, if a vector $z$ is not in the span of the set of "basis" vectors, by linear independence, a composite set, $S\cup{\{z\}}$, is linearly independent, and so is a basis.

Now at this point, in the book, they haven't yet proven any results about dimension, but it's intuitively clear that the theorem is true.

Question: Some of the basic results in LA are intuitively obvious, e.g. they are immediate from the definitions and axioms given (plus, implicitly, some other results about the properties of the underlying field and certain set-theoretic properties of the vector space itself). And, yet when we get to the proof, we start this constructive schtick that really starts confusing the hell out of me, and I start losing the clearer picture I had from the definition of the terms. So, at the level of the above mentioned theorem, is the explanation I outline count as a sufficient proof? or, do I need (i.e. is it always better) to spell it all out constructively?

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Your proof sketch is confusing due to lack of specificity about bases. The basis in the first line is a basis for $U$, while the goal (which is suggested although not achieved in the last line) is finding a basis for $V$. Without specifying this sort of thing, you haven't proved much. –  Tabes Bridges Jan 16 '13 at 18:02
    
What you really ought to say is that any set of linearly independent vectors can be expanded to a basis. Your proof doesn't quite work - you need to use that $\mathbb R$ is a field somewhere. Note that if you take the set of integer vectors, some of these "obvious" theorems are no longer true - you can have a set of linear independent integer vectors so that it cannot be completed to a linear integer basis for the integer vectors. On the other hand, if you take vectors with rational coordinates, you still hve all these nice properties. –  Thomas Andrews Jan 16 '13 at 18:03
    
To expand on @TabesBridges, there are infinite-dimensional vector spaces, and it turns out you need to use the axiom of choice to prove that arbitrary infinite-dimensional vector spaces have a bases. –  Thomas Andrews Jan 16 '13 at 18:04
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Your "proof" isn't one. You didn't make use of the fact that $\{x_1,x_2,\dotsc,x_{n-1}\}$ has dimension one less than $V$, but you need to, since the theorem wouldn't be true without that. The error is in the last sentence; the fact that $S\cup\{z\}$ is linearly independent doesn't imply that it's a basis. –  joriki Jan 16 '13 at 18:07
    
@Thomas: The theorem is about a vector space; integer "vectors" don't form a vector space. Also, $\mathbb R$ isn't mentioned anywhere. –  joriki Jan 16 '13 at 18:08

1 Answer 1

up vote 2 down vote accepted

At heart, you appear to be confusing "being a basis for $V$" with "being a basis for some subspace of $V$." Any linear independent set of vectors is a basis for some subspace of $V$, but that doesn't mean that any linear independent set of vectors is a basis for $V$.

The fact:

If a finite basis for a vector space $V$ exists, then any other linearly independent set of the same size as the basis is also a basis.

is something that you have to prove, although that may seem intuitively obvious. Basically, this means that the "dimension" of a (finite-dimensonal) vector space is well-defined. Indeed. the theorem you are trying to prove can be seen as the first lemma towards proving the above fact.

Essentially, your effort has merely shown that $x_1,\dots,x_{n-1},z$ are linearly independent. You haven't shown that they span all of $V$. Things you have to work with to prove this fact are that $x_1,\dots,x_n$ span $V$ and $z$ is not in the subspace spanned by $x_1,...,x_{n-1}$.

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