# min & max number of linearly independent vectors required to span $F^{n}$

What is the minimum and maximum number of linearly independent vectors required to span $$F^{n}$$?

I'm going to guess you need exactly $$n$$ vectors to span the entire space of $$F^{n}$$?

No more than $$n$$, and no less than $$n$$, but exactly $$n$$.

True or false? and Why?

• Welcome to stackexchange. Yes, exactly $n$. That is a standard theorem in linear algebra. It;s proved in all the books and courses. This website isn't really the place to ask "why". – Ethan Bolker Feb 23 at 14:21
• apparently every book except "linear algebra demystified" which makes you guess that its a theorem... – John Proxer Feb 23 at 14:46

If you were to construct a linearly independent set, you'd notice that it cannot contain more elements than $$dim(\mathbf{F}^n)=n$$. Also, any set S of $$n$$ linearly independent vectors in $$\mathbf F^n$$ will be a basis for $$\mathbf{F}^n$$, hence S spans $$\mathbf{F}^n$$. Adding arbitrarily many more vectors to S does not change the fact that it spans $$\mathbf{F}^n$$ because these added vectors were already in the span of S.
And the answer to your question is this: you need $$n$$ vectors to form a linearly independent set S which spans the whole vector space. Any more vectors added to S, and S is no longer linearly independent, since it already spanned the whole vector space.