In Kreyzig's Functional Analysis with Applications, the dimension of a finite-dimensional vector space is defined to be $\dim(X) = n$ if there exists a linearly independent set of $n$ vectors in $X$ and no set of $n+1$ vectors is L.I.

I'm trying to prove that for all $n$ the following are equivalent, so that dimension becomes a well-defined value:

  1. $\dim(X) = n$.
  2. $X$ has a Hamel basis (an L.I. spanning set), and every L.I. set of $n$ vectors in $X$ is a basis.
  3. There exists a spanning set of $n$ vectors, and no set of $n-1$ vectors spans $X$.

So far I've got (1) and (2) are equivalent because I've shown every maximal L.I. set is a basis, and I've also got that (3) implies a basis exists (because minimal spanning sets are such), but I'm having trouble showing (2) implies (3) and finishing proving (3) implies (2).

Separately I also have that the representation of a vector $x$ as a linear combination of the basis vectors is unique (but I'm not sure if that's useful here).

How should I proceed from here?


1 Answer 1


For 3 implies 2, consider a l.i. list of $n$ vectors $u_1,...,u_n$ and a Hamel basis (the one you showed exists) $v_1,...,v_n$. You can make a list of $n+1$ vectors $u_1,v_1,...,v_n$ which will necessarily be l.d.; you can throw away one of the $v_j$ and still have a spanning set. Iterate to show that the $u_j$ are also a spanning set.

For 2 implies 3, you already have a spanning set of $n$ vectors, so you only have to show no set of $n-1$ vectors spans the space. I have a vague idea that you can force a contradiction by representing the hamel basis of $n$ vectors in terms of the $m < n$ l.i. subset of the spanning set of $n-1$ vectors. I mean, you'd have

$$a_1u_1 + ... + a_nu_n$$

You'd have some degrees of freedom in the choice of the $a_i$ so as to make the original hamel basis l.d.


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