# Proof Verification: Prove V is infinite-dimensional iff there is a sequence of vectors in V that is linearly independent for every positive integer m

I am currently self-studying Axler's Linear Algebra Done Right. I came up with a proof for the question listed in the title, but I want to check if my approach is correct. I looked at Question about my approach to linear algebra proof, Axler Ch.2A #14 which offers helpful suggestions, but takes a slightly different direction than my proof. Here is my proof:

$$\triangle$$ $$Proof:$$

Assume V is an infinite-dimensional vector space. To prove that there is a sequence of vectors such that $$v_1, v_2, ..., v_m$$ is linearly independent for every $$m \in \mathbb{N}$$, I will utilize induction. For m=1, just choose any non-zero vector, and you have a linearly independent list (I proved this in an earlier problem, so I will omit the proof here). Now, before going to the inductive step, consider the following lemma

$$\textit{Lemma: Given a list of vectors v_1,v_2,...,v_n in an infinite-dimensional vector space,}$$ $$\textit{V, there exists a vector w \in V such that w \notin span(v_1,v_2,...,v_n)}$$.

To prove this, assume for contradiction that there is a list of vectors $$v_1,v_2,...,v_n$$ in V such that for every arbitrary vector $$w\in V$$, $$w\in span(v_1,v_2,...,v_n)$$. Since vector spaces are closed under addition and scalar multiplication, $$span(v_1,v_2,...,v_n) \subseteq V$$. However, since we assumed $$w\in V$$ and $$w\in span(v_1,v_2,...,v_n)$$, we also have that $$V \subseteq span(v_1,v_2,...,v_n)$$. Thus, $$V=span(v_1,v_2,...,v_n)$$. This is a contradiciton since we assumed $$V$$ was infinite dimensional, and therefore it cannot be spanned by a list of vectors. Thus, there must exist some $$w\in V$$ such that $$w \notin span(v_1,v_2,...,v_n)$$.

Now, apply the inductive step. Assume that we have a list of $$m-1$$ linearly independent vectors $$v_1,v_2,..,v_{m-1}$$ where $$m \in \mathbb{N}$$. We can apply our lemma which tells us that there must exist $$v_m \in V$$ such that $$v_m \notin span(v_1,v_2,..,v_{m-1})$$. This tells us that the list $$v_1,v_2,...,v_{m-1},v_m$$ is linearly independent (Note that this should be proven, but I did so in an earlier question, so I take the result as given). Thus, we have proved the first direction of our statement.

Now, for the other direction. Assume that there is a sequence of vectors $$v_1,v_2,...,v_m$$ in V that is linearly independent for every $$m \in \mathbb{N}$$. For contradiction, assume that V is finite dimensional. This implies that there is a list of vectors of length $$x\in \mathbb{N}$$ that spans V. However, by our initial assumption, we have that there exists a sequence of vectors in V of length $$x+1$$ where each vector is linearly independent. This is a contradiction since a linearly independent list of vectors cannot have greater length than a spanning list of vectors. Thus, V must be infinite dimensional.

Thanks for taking the time to read. Any suggestions or corrections are much appreciated!

• Your argument works but in my opinion makes unnecessary use of proofs by contradiction. Using the other facts you already know, it's easier to argue that if $V$ is finite-dimensional (i.e. spanned by a finite list), then there can't be arbitrarily long linearly independent lists; and conversely if there aren't arbitrarily long linearly independent lists, then a maximally long one must span, so $V$ is finite-dimensional. Oct 23, 2023 at 0:25