I noticed that the definition of the dimension of a finite-dimensional vector space relies on the fact:

If a vector space V has a finite basis, then every basis of V is finite.

Can someone tell me how to prove it? Thanks.


Hint Let $B=\{ b_1,.., b_n \}$ be a fixed finite basis.

We can prove something stronger:

Lemma If a set has at least $n+1$ elements, it is linearly dependent.

Proof: Let $v_1,..,v_{n+1} \in S$. Write $$v_i=\sum_{j=1}^n c_{ij}b_j$$

Then, replacing each $v_i$ by the above expression,
$$x_1v_1+...+x_{n+1}v_{n+1} =0$$ leads to an homogeneous system of $n$ equations with $n+1$ unknowns (the coefficient matrix being exactly $C^T$). This homogeneous system has more unknowns that variables, and hence as non-trivial solution.

This proves the linear dependence.

Corollary Any other basis has exactly $n$ elements.

Proof: If $B'$ is any other basis, by Lemma $B'$ has at most $n$ elements. Interchanging $B$ and $B'$ you get also that $B'$ has at least $n$ elements.


A basis is by definition a set of linearly independent vectors which generate $V$. If a basis of V is finite, than if an infinite basis exists every vector of this basis could be written with a linear combination of vectors of the finite basis, so the infinite vectors are all linearly dependent from the vectors of the "first" basis (that is finite by hypothesis).

  • $\begingroup$ If you define dimension in terms of the size of a basis, how can you use that isomorphisms should preserve dimension, to show that dimension is well defined? To prove that isomrphisms prove dimension, you need to know it is well defined first. $\endgroup$ – James Oct 15 '16 at 16:24

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