# Some questions on basic linear algebra about dimension and basis

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?

• 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). – Morgan Rodgers Nov 12 '19 at 19:39
• For question 2, if $\dim{W} = \infty$ that would be bigger than $\dim{V}$. – Morgan Rodgers Nov 12 '19 at 19:41
• Do you have the definition "A vector space is $n$-dimensional if it has a basis with $n$ elements"? – 79037662 Nov 12 '19 at 19:42
• @MorganRodgers, to be honest I didn't understand your first comment. Could you explain it one more time? – ZFR Nov 12 '19 at 19:59
• 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 $n$. – Morgan Rodgers Nov 12 '19 at 20:02

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.

• Regarding question 1: You said "by your second definition.." but what if $W$ is infinite-dimensional? – ZFR Nov 12 '19 at 20:43
• Yes, I was editing it! haha thanks! – IamWill Nov 12 '19 at 20:44
• Regarding question 2: You have shown that $\dim W\leq m$ but how it follows that $W$ is finite-dimensional by these definitions? – ZFR Nov 12 '19 at 20:48
• Because every basis of $W$ must have at most $m$ vectors, and then you apply your second definition again. – IamWill Nov 12 '19 at 20:51
• 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. – ZFR Nov 12 '19 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 $$n \le p$$.