In Sheldon Axler's "Linear Algebra Done Right" 3rd edtion Page 36 he worte:Proof of every subspaces of a finite-dimensional vector space is finite-dimensional

The question is: I do not understand the last sentence"Thus the process eventually terminates, which means that U is finite-dimensional". So far I can understand that the subspace $U$ has limited length of independent vector list, but how to reach from here to the conclusion that subspace $U$ is finite dimensional? By 2.23 we see that the length of spanning list of vectors can be more than(rather than less than) the finite list vetors' length. Thus finite length of independent vectors list does not imply finite length of spanning list!

My proof: By the definition of subspace, we see that $U$ is a subset of $V$(with some other restrictions). Then, for every $u\in U$, $u\in V$, implying $u\in span(v_1,v_2,...,v_m)$ for $V= span (v_1,v_2,...,v_m)$ (Here I used the difinition 2.10). QED.

2.23 Length of linearly independent list

length of spanning list In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

2.10 Definition finite-dimensional vectors pace

A vector space is called finite-dimensional if some list of vectors in it spans the space.

Please verify why Professor Axler's proof is valid and point out any mistakes I've made in my understanding or proof if there are some, thanks in advance!

  • $\begingroup$ Your "proof" is one of the most common and painful mistakes many, many students commit: you must prove that there's a finite generating set of vectors in $\;U\;$ ... $\endgroup$
    – DonAntonio
    Apr 22, 2017 at 11:01
  • $\begingroup$ By the way, is a very poor decision, imo, in that book to denote the inductively built set of generators of $\;U\;$ as $\;v_1,v_2,...\;$ , instead of the probably clearer $\;u_1,u_2,...\;$ . $\endgroup$
    – DonAntonio
    Apr 22, 2017 at 11:03
  • 2
    $\begingroup$ That "The process eventually terminates" is because the list so built is of vectors in $\;U\subset V\;$ , so they are linearly independent and in $\;V\;$, and since $\;V\;$ is finite dimensional...etc. $\endgroup$
    – DonAntonio
    Apr 22, 2017 at 11:05
  • 2
    $\begingroup$ This seemes easier: if $V$ is finite-dimensional, say $\mbox{dim}\,V=n$, then $n+1$ vectors in $V$ are always linearly dependent. Take $n+1$ vectors in $U$, they are also in $V$ so... But then the dimension of $U$ can be at most... $\endgroup$
    – StackTD
    Apr 22, 2017 at 11:16
  • $\begingroup$ You have proved that if V is finite dimensional and U is a subset of V then U is the subset of a finite-dimensional space. In other words, you have proved that the hypotheses of the theorem imply the hypotheses of the theorem, which is not terribly useful. Note that you have proved that U is contained in the span of finitely many vectors (which is essentially just your assumption), not that it is the span of finitely many vectors. $\endgroup$ Apr 22, 2017 at 13:44

1 Answer 1


The proof in the book is correct and yours isn't.

Indeed, it cannot be proved that a subspace of a finitely generated vector space $V$ is finitely generate only on the basis of the vector space axioms, without using a very specific property of the field of scalars, namely that it is a field, so a noetherian ring.

It is true that any $u\in U$ is a linear combination of a spanning set for $V$, but what you need is a finite set of vectors in $U$ (not just in $V$) that spans $U$.

The key points in the proof are:

  1. no linearly independent set of vectors can have more elements than the dimension of the space $V$;

  2. if $(v_1,\dots,v_{j-1})$ is linearly independent and $v_j\notin\operatorname{span}(v_1,\dots,v_{j-1})$, then also $(v_1,\dots,v_{j-1},v_j)$ is linearly independent.

If $U$ were not finitely generated, then the process outlined in the proof, which uses the second key point, would not stop, contradicting the first key point.

Note that the first key point relies on the fact that nonzero scalars have an inverse. Without this property, one cannot prove it.

  • $\begingroup$ Thanks for pointing out my mistakes. So, you mean that in the definition 2.10,"in it" CANNOT be omitted, right? Alright, the symbol $v$ actually denotes vector in $U$, which is what I neglect. Thank everyone for giving me such a clear explanation! $\endgroup$
    – Frank Li
    Apr 22, 2017 at 20:03
  • $\begingroup$ @FrankLi I don't have the book, so I don't know what definition 2.10 says. $\endgroup$
    – egreg
    Apr 22, 2017 at 20:06
  • $\begingroup$ 2.10 Definition finite-dimensional vectors pace A vector space is called finite-dimensional if some list of vectors in it spans the space. (As posted in my question^_^) $\endgroup$
    – Frank Li
    Apr 23, 2017 at 2:17
  • $\begingroup$ See math.stackexchange.com/a/2865790/268333 for an expansion of the last point. $\endgroup$
    – tparker
    Jul 29, 2018 at 5:22

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