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?

  • 1
    $\begingroup$ 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". $\endgroup$ – Ethan Bolker Feb 23 at 14:21
  • $\begingroup$ apparently every book except "linear algebra demystified" which makes you guess that its a theorem... $\endgroup$ – 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.

I've found this helpful: any spanning set can be reduced into a basis, every linearly independent set can be grown into a basis and finally adding elements to a basis set does not change the fact that it spans the whole vector space, but it does make it a linearly dependent set since the added vector is in the span of the original n vectors in the basis.

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.

Hope that answers your question.


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