# Prove every subspace of a finite-dimensional vector space is finite-dimensional.

Similar question But don't directly solver my confusions.

In Linear Algebra Done Right, it said:

Proof:

Step 1: If $$U = \{0\}$$, then $$U$$ is finite-dimensional and we are done. If $$U \neq \{0\}$$, then choose a nonzero vector $$v_1 \in U$$.

Step J: If $$U = span(v_1,...,v_{j-1})$$, then $$U$$ is finite-dimensional and we are done. If $$U \neq span(v_1,...,v_{j-1})$$, then choose a vector $$v_j \in U$$ such that $$v_j \not \in span(v_1,...,v_{j-1})$$.

After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is the span of the previous vectors. Thus, after each step we have constructed a linearly independent list, by Linear Dependence Lemma. This linearly independent list cannot be longer than any spanning list of $$V$$. Thus the process eventually terminates, which means $$U$$ is finite-dimensional.

I have problem on "this linearly independent list cannot be longer than any spanning list of $$V$$." Should it be $$U$$?

If it is $$U$$, how do we know the spanning list of $$U$$ is finite? Or is it the definition, because I remember in Chapter 1, it said list should have finite finite length.

• You can replace the word "list" by "set" if that clarifies things. – user370967 Jan 7 '19 at 23:33
• Perhaps a simpler (depending on the machinery available) proof would be: Take a basis of $U$. This is a linear independent family of $V$. Hence it can be extended to a basis of $V$. As the dimension is the length of a basis, we conclude $\dim U\le \dim V<\infty$. Then again, the needed machinery should be available to you when you know the concept of dimension of vector space ... – Hagen von Eitzen Jan 8 '19 at 11:50
• @Math_QED Actually, it is my opinion that "set" instead of "list" (or even better: "family") de-clarifies things in the context of linear independence, bases, etc. For example, if $A$ is a singular square matrix, the list of its column vectors is always a linearly dependent list, but the set of its column vectors might be a linearly independent set - a highly undesireable situation. – Hagen von Eitzen Jan 8 '19 at 11:53

It is supposed to be $$V$$. The proof constructs linearly independent $$v_1, v_2, \ldots$$, which are going to be linearly independent regardless of whether you consider them in $$U$$ or $$V$$.
Because they're linearly independent as vectors in $$V$$, the process must terminate eventually. It won't always terminate after $$\operatorname{dim} V$$ vectors (that would imply $$U = V$$), but there is always this fixed upper bound $$\operatorname{dim} V$$ that is independent of $$U$$. That should answer your second question.
• so it means the constructed sequence $v_1, v_2,...$ lies in $V$ which is independent in $U$ or $V$. since $V$ is finite-dimensional, the linear independent list of $V$ is also finite. Therefore the process will terminate. – JOHN Jan 9 '19 at 23:33
No, it should be $$V$$ as written.
The point is that the constructed sequence $$v_1,v_2,\dots$$ lies in $$V$$, and is linearly independent by construction.