# Proof that every subspace of a finite dimensional vector space

In Axler's Linear Algebra Done Right, Axler proves this statement by the following:

From what I can understand, he constructs a list of vectors in $$U$$ such that they are all linearly independent and since the length of linearly independent list cannot be greater than the length of the spanning list, the linearly independent list is finite thus proving U is finite dimensional.

My question: How does this idea of being able to construct a finite lineally independent list in $$U$$ show that $$U$$ is finite dimensional?

My thinking: For my other questions on this site, I usually include how I thought about it however for this I really don't know/have any ideas. I feel like I might be misinterpreting his proof somehow. I've seen the answer to this question Prove that "Every subspaces of a finite-dimensional vector space is finite-dimensional" however I still can't answer my question.

• I think what you might be missing is understanding step $j$. There Axler compares $U$ with the span of the list of vectors. In short, you don't only construct a linearly independent list in $U$ but instead constructing a linearly independent list that --spans-- $U$. Jul 3, 2022 at 15:50
• "How does this idea of being able to construct a finite lineally independent list in 𝑈 show that 𝑈 is finite dimensional?" No, some finite linearly independent list is not enough to show that $U$ is finite dimensional. Instead, he has shown that for any $n\in\mathbb{N}$ there is a linearly independent list of length $n$ in $U$. Since this gives an infinite, linearly independent list of vectors in $U$, and hence in $V$, it contradicts the premise that $V$ is finite dimensional. Jul 3, 2022 at 15:58
• The author is adding vectors of $U$ to the spanning list as long as the list doesn't span $U$. He then argues that this process of growing the list of vectors that span $U$ must terminate because the entire vector space is spanned by only $n$ vectors. He's essentially saying that we can construct a basis for any subspace $U$ and since it takes only $n$ vectors to span the parent space, it can take no more than $n$ vectors to span a subspace. Jul 3, 2022 at 16:17

For finite-dimensional vector spaces Axler proved that if $$(v_1,\ldots,v_j)$$ is any linearly independent list and $$(w_1,\ldots,w_n)$$ is any spanning list, then $$j \le n$$.
Now let us consider a finite-dimensional vector space $$V$$ with a finite spanning list of length $$n$$ and a subspace $$U$$ of $$V$$.
The point in Axler's proof is not that we are able to construct some finite linearly independent list in $$U$$, but that any finite linearly independent list of any length $$j$$ which is not a spanning list can be prolongated to a linearly independent list of length $$j+1$$.
Therefore if $$U$$ would not be finite dimensional, then it would not have a finite spanning list and therefore we could construct linearly independent lists in $$U$$ of arbitrary length, in particular one of length $$n+1$$. This list is also linearly independent in $$V$$ which contradicts the fact the length of any linearly independent list in $$V$$ can at most be $$n$$.