# Another approach for dimension of a vector space

I am preparing a lecture note for a primary course on Vector Spaces and I am developing basis and linearly independent sets. There I took the following path:

1. Define the linear span $$L(S)$$ of a subset $$S$$ of a vector space;
2. Both the sets $$S_1=\{(1,0),(0,1)\}$$ and $$S_2=\{(x,x+1),x\in\mathbb{R}\}$$ have the same span $$\mathbb{R}^2$$;
3. We can delete infinitely many elements from $$S_2$$ and still retain the same span but no point can be deleted from $$S_1$$ in order to retain the same span;
4. Define a basis by: $$S$$ is a basis if $$S$$ spans the space $$V$$ and no proper subset of $$S$$ spans $$V$$;
5. Assume the existence of a basis for any vector space;
6. Define a linearly independent set by: $$S$$ is linearly independent if $$\forall \alpha\in S$$, $$\alpha\notin L(S\setminus\{\alpha\})$$;
7. Define a maximal linearly independent set by: $$S$$ is a maximal linearly independent set if for every superset $$S'\supset S$$, $$S'$$ is not linearly independent;
8. Show that $$S$$ is a basis iff $$S$$ is a maximal linearly independent set;
9. Existence of a maximal linearly independent set is guaranteed by the assumed existence of a basis.

After this, I want to define the dimension of a vector space. For that, I need to show that every basis or every maximal linearly independent set in $$V$$ has the same cardinality. And I am stuck to prove this.

A few things to mention:

• This is a primary course on vector spaces and the existence theorem for a basis (and the Zorn's lemma) is not there in their syllabus.
• I am eventually going to move to finite dimensional spaces, but only after defining dimension in the general set up.

• This link might be useful to you math.stackexchange.com/questions/86762/… – sathishT May 15 '20 at 8:17
• I appreciate the flow of your path to teach vector spaces. However i suppose it will be easier for students to move from finite dimensional vector spaces to infinite dimensional spaces. Other way around seems quite tricky – sathishT May 15 '20 at 8:21
• I agree with you. But actually I find this approach to be more natural. That is why I am curious to consult some books. But if I don't get an answer, maybe I will have to impose a jump cut to finitely generated spaces and develop the concept of dimensions there. – Subhajit Paul May 15 '20 at 8:53

The approach that I've been taught with, which I think is more conventional, is to first learn spans, then linear independence, then define a basis and show that all bases have the same cardinality (at least in a finite dimensional case. I'm not so sure about infinite dimensional cases).

To prove the statement that every basis has the same cardinality, I'll make use of theorems on matrices and theire row-reduced echelon forms.

Lemma 1: If $$B=\{v_1, v_2, \dots, v_n\}$$ is a basis for a vector space $$V$$, then any set of vectors in $$V$$ with more than $$n$$ elements is linearly dependent.

Proof

let $$S$$ be a subset of $$V$$ with more thatn $$n$$ elements. in particular, let $$\alpha_1, \alpha_2, \dots, \alpha_m$$ be distinct vectors in $$S$$ with $$m > n$$.

$$c_1\alpha_1 + c_2\alpha_2 + \dots + c_m\alpha_m = 0$$

Converting everything to coordinates of the basis, we get

$$c_1[\alpha_1]_B + c_2[\alpha_2]_B + \dots + c_m[\alpha_m]_B = 0$$

But $$[\alpha_i]_B$$ are vectors in $$\mathbb F^n$$ where $$\mathbb F$$ is the field of the vector space (you can take it to be $$\mathbb R$$ if you haven't learnt fields).

If we form the matrix $$\left( [\alpha_1]_B \;\; [\alpha_2]_B \;\; \dots \;\;[\alpha_m]_B\right)$$, then this is a $$n\times m$$ matrix with $$n < m$$, so it's row-reduced echolon form will certainly have a non-pivot column and thus there are non trivial solutions for $$c_1, c_2, \dots, c_m$$. $$\;\;\blacksquare$$

Lemma 2: If $$B=\{v_1, v_2, \dots, v_n\}$$ is a basis for a vector space $$V$$, then any set of vectors in $$V$$ with less than $$n$$ elements does not span $$V$$.

Proof

By following the same procedure in the previous example, you'll get an $$n\times m$$ matrix where $$n > m$$, so it's row-reduced echelon form will have a zero row, let's say in row $$i$$. Then a vector whose $$i^{th}$$ coordinate is not zero, such as $$v_i$$, will not be in the span of this set. $$\;\;\blacksquare$$

By combining these two lemmas, every basis must therefore have the same cardinality.

• This approach I also learnt as a student. But somehow I found it lacks motivation. That is why I came across with the other approach. If I don't find an answer, maybe I will impose a jump cut to finitely generated spaces and try to develop dimension. – Subhajit Paul May 15 '20 at 8:49
• Actually the textbook I use talks about finite and infinite-dimensional spaces together. (Hoffman and Kunze if you're interested). However, I think dimension becomes a very tricky thing in infinite-dimensional spaces, and the discussion of it should be restricted to finite dimensional spaces in an introductory course – Saad Haider May 15 '20 at 8:54
• Yes. I also followed Hoffman and Kunze which lead me to define linear span of an infinite set at this entry-level lectures. At this point, I am just looking for a bijection (or guarantee of the existence of a bijection) between any two bases of a vector space. – Subhajit Paul May 15 '20 at 8:59