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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.

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  • $\begingroup$ You can replace the word "list" by "set" if that clarifies things. $\endgroup$ – user370967 Jan 7 '19 at 23:33
  • $\begingroup$ 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 ... $\endgroup$ – Hagen von Eitzen Jan 8 '19 at 11:50
  • $\begingroup$ @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. $\endgroup$ – Hagen von Eitzen Jan 8 '19 at 11:53
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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.

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  • $\begingroup$ 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. $\endgroup$ – JOHN Jan 9 '19 at 23:33
  • $\begingroup$ @JOHN Yes, precisely. $\endgroup$ – Theo Bendit Jan 10 '19 at 1:24
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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.

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