I am reading some lectures on linear algebra which are on Russian language and few moments have confused me and I was trying to understand them correctly but I failed to do it.

Firstly, let me give you some preliminary definitions and theorems from my lectures:

Definition: A vector space $V$ is called finite-dimensional, if it has a basis consisting of finitely many vectors. Otherwise, we call the space to be infinite-dimensional.

Theorem: In finite-dimensional vector space each basis has the same number of vectors.

The proof of this theorem is based on the following lemma:

Lemma: Let $\{e_1,\dots,e_m\}$ and $\{f_1,\dots,f_n\}$ be two linearly independent system of vectors such that the second system is contained in the linear span of the first. Then $n\leq m$.

Definition: The dimension of finite-dimensional vector space $V$ is the number of elements in each basis of $V$. If $V$ is infinite-dimensional, then we write $\dim V =\infty$.

Then he is proving the following statement which confuses me.

Statement: Subspace $W$ of finite-dimensional space $V$ is finite-dimensional and $\dim W\leq \dim V$.

Proof: Since $V$ is finite-dimensional then $\dim V=m$ and let $\{e_1,\dots,e_m\}$ basis of $V$. Suppose that $\dim W>m$ then $W$ contains linearly independent vectors $f_1,\dots,f_n$ with $n>m$. Then $\{f_1,\dots,f_n\}\subset \langle e_1,\dots,e_m\rangle=V.$ But this contradicts to the above lemma. Hence $\dim W\leq \dim V$.

I understood the idea of the proof but cannot understand some technical moments.

Question 1: If $\dim W>m$ then why $W$ contains linearly independent vectors $f_1,\dots,f_n$ with $n>m$. Intuitively I know this but can anyone show it rigorously?

Question 2 (sorry for stupid question): Suppose we have shown that $\dim W\leq \dim V$ then how it follows that $W$ is also finite-dimensional?

  • $\begingroup$ If there were not $n$ linearly independent vectors in $W$, then you would take as many as you could and they would have to span $W$ (i.e. would be a basis with less than or equal to $m$ elements). $\endgroup$
    – xxxxxxxxx
    Nov 12, 2019 at 19:39
  • $\begingroup$ For question 2, if $\dim{W} = \infty$ that would be bigger than $\dim{V}$. $\endgroup$
    – xxxxxxxxx
    Nov 12, 2019 at 19:41
  • $\begingroup$ Do you have the definition "A vector space is $n$-dimensional if it has a basis with $n$ elements"? $\endgroup$
    – 79037662
    Nov 12, 2019 at 19:42
  • $\begingroup$ @MorganRodgers, to be honest I didn't understand your first comment. Could you explain it one more time? $\endgroup$
    – RFZ
    Nov 12, 2019 at 19:59
  • $\begingroup$ Sure, if you take a collection of linearly independent vectors in $W$, starting with a single vector, then at each step either 1: the collection you have spans all of $W$, and so is a basis for $W$ or 2: they don't span all of $W$ so you can add a vector to get a larger collection of linearly independent vectors. You just do this until you have more than $n$. $\endgroup$
    – xxxxxxxxx
    Nov 12, 2019 at 20:02

2 Answers 2


Question 1: Suppose $\dim W > m$. By your second definition, the dimension of a finite-dimensional vector space is the number of elements of each basis of this vector space. This surelly works for vector subspaces too. So, if $W$ is finite dimensional, each basis of $W$ has more than $m$ elements. But we know that the elements of a basis must be linearly independent, by the definition of a basis. So, we conclude that $W$ contains linearly independent vectors $f_{1},...,f_{n}$, for some $n>m$. If $W$ is infinite-dimensional, the statement still holds. To see this, suppose that there is no set of linearly independent vectors $f_{1},...,f_{n}$ in $W$. Then, every set of $n$ vectors is linearly dependent, and thus $\dim W < n$, which contradicts the fact that $W$ is infinite-dimensional.

Question 2: If $\dim W \le \dim V$ then $\dim W \le m$, since $m = \dim V$. Thus, the dimension of $W$ must be, at most, equal to $m$, which is a finite number. It follows that $W$ is finite-dimensional, once its dimension if finite.

  • $\begingroup$ Regarding question 1: You said "by your second definition.." but what if $W$ is infinite-dimensional? $\endgroup$
    – RFZ
    Nov 12, 2019 at 20:43
  • $\begingroup$ Yes, I was editing it! haha thanks! $\endgroup$
    – IamWill
    Nov 12, 2019 at 20:44
  • $\begingroup$ Regarding question 2: You have shown that $\dim W\leq m$ but how it follows that $W$ is finite-dimensional by these definitions? $\endgroup$
    – RFZ
    Nov 12, 2019 at 20:48
  • $\begingroup$ Because every basis of $W$ must have at most $m$ vectors, and then you apply your second definition again. $\endgroup$
    – IamWill
    Nov 12, 2019 at 20:51
  • $\begingroup$ It is still not so clear to me. You said that each basis of $W$ must have at most $m$ vectors. Why? What if some basis has $>m$ vectors. $\endgroup$
    – RFZ
    Nov 12, 2019 at 20:57

By definition, if $n=\dim W$ then $n$ is the number of elements in (any) a basis of $W$.

Suppose $W \subset V$ and $p=\dim V $ is finite. Let $n=\dim W$, then there is a basis $b_1,...,b_n$ for $W$.

If $V \setminus W$ is non empty we can find some $b_{n+1} \in V \setminus W$ and the resulting collection $b_1,...,b_{n+1}$ is linearly independent.

We can continue this process (with $\operatorname{sp} \{b_1,...,b_{n+1} \}$ in place of $W$) a finite number of times. When it terminates, the resulting collection will be a basis of $V$ from which it follows that $n \le p$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .