# $V$ has a finite spanning set $\implies V$ is finite-dimensional

Let $V$ be a vector space over the field $F$. Suppose there are a finite number of vectors $\alpha_1, . . . , \alpha_r$ in $V$ which span $V$. Prove that $V$ is finite-dimensional.

Does it suffice to just say that no linearly independent set of $V$ can have more than $r$ elements. So a basis for $V$ must have $r$ or less elements in it. And hence $V$ has a finite basis?

Or am I missing/ignoring something??

• If you have a proof that no linearly independent set in $V$ can contain more than $r$ elements, that's enough. Otherwise, you could argue with the fact that every spanning set contains a basis (might also require a proof, but you can restrict yourself to proving it for finite spanning sets here). – Daniel Fischer May 30 '14 at 16:19
• @Daniel Fischer: Yes. I have a proof for the former. Thanks. Problem seemed too easy that's all. – Ishfaaq May 30 '14 at 16:21
• @Daniel Fischer: One more thing if it's alright. To prove your second suggestion - Consider a finite spanning set $A$ for a vector space. Consider its largest linearly independent subset, $B = \{b_1, b_2, ..., b_n\}$. Suppose there is $a \in A \setminus B$ which is not a linear combination of vectors in $B$. Then suppose $k_1b_1 + k_2b_2 + ...+ k_nb_n + ka = 0$. It can be argued that $k$ must be $0$ and hence all the scalars $k_i = 0$ contradicting the maximality of $B$. Hence $B$ can be proved to be a basis. Sound good enough?? – Ishfaaq May 30 '14 at 16:35
• "It can be argued" is not a phrase that fits particularly well in a proof. For a proof, you need to spell that argument out (it's after all the crucial point). Then it works. But it's IMO easier/nicer to use a minimal spanning set contained in $A$, and show that that is linearly independent. – Daniel Fischer May 30 '14 at 16:43
• @Daniel Fischer: Too long for a comment though. And you're right. That is more concise. Thanks loads for the input. Much appreciated. – Ishfaaq May 30 '14 at 16:46

Suppose the vectors $\alpha_1, ... , \alpha_r$ form a basis for $V$. ie. They are linearly independent and span $V$. Then by definition $V$ is finite dimensional. So instead assume that for some $\alpha_i$ there exists scalars $c_1, ... , c_{r-1}$ in $F$ such that (without loss of generality let $\alpha_i$ be moved to become the rth element)
$c_1\alpha_1 + ... + c_{r-1}\alpha_{r-1} = \alpha_i$ Then we can remove $\alpha_i$ from $V$ and form a new subset of vectors $\alpha_1, ... , \alpha_{r-1}$ that clearly still span $V$.
If the vectors $\alpha_1, ... , \alpha_{r-1}$ are linearly independent then they form a basis for $V$ and again by definition $V$ is finite dimensional. If the vectors $\alpha_1, ... , \alpha_{r-1}$ are linearly dependent then we repeat this process until we form a basis for $V$ and hence prove our theory that $V$ is finitely dimensional.