# Dimension of a infinite dimensional vector space

I have a simple multiple choice I've been trying to understand here for a while now.

Suppose that $$F$$ is spanned by the functions $$f_1,f_2,f_3,f_4$$. What can we conclude about $$\text{dim}(V)$$?

a) $$\text{dim}(V)=4$$

b) $$\text{dim}(V)\geq 4$$, but we know nothing else

c) $$\text{dim}(V)\leq 4$$, but we know nothing else

d) We can conclude nothing about $$\text{dim}(V)$$

e) None of the above

NORMALLY I would say that the answer is $$b)$$ but that only holds for a finite dimensional vector space. I'm not sure if I can assume that $$b$$ is true if $$F$$ is infinite dimensional.

Can someone guide me on this? Thanks!

• The answer is c). There is no infinite dimensional space in this question. – Kavi Rama Murthy Oct 19 '20 at 5:56
• Okay so I need to ask. Why? I thought since F could be anything, it's infinite dimensional. Also why is it c? For that to happen, don't the vectors have to all be linearly independent? – Future Math person Oct 19 '20 at 5:58

## 1 Answer

If $$f_1,f_2,f_3,f_4$$ are linearly independent, then $$\{f_1,f_2,f_3,f_4\}$$ is a basis of $$F$$, hence $$\dim (V)=4.$$

If $$f_1,f_2,f_3,f_4$$ are linearly dependent, then $$\dim(V) \le 3.$$

The answer is c).

• How do you know the vectors are linearly dependent? Is it because they span $F$ so it's possible to express one or more vectors as a linear combination of each other and hence the dimension can be at most $3$? – Future Math person Oct 19 '20 at 6:04