# Spanning of a Vector Space

Prove or disprove: If a set of $$n$$ vectors in the vector space $$V$$ is linearly independent, then the set of $$n-1$$ vectors cannot span $$V$$.

I believe the statement to be false. Can't there be a line, plane, or hyperplane which spans a vector space? Each having their own set of linearly independent vectors?

$$n-1$$ vectors can span at maximum a $$n-1$$ dimensional space. So if your space is of dimension $$n$$, $$n-1$$ vectors cannot span it.
A vector space $$V$$ with dimension $$n$$ has, by definition, a basis of $$n$$ linearly indipendent vectors which span $$V$$. It is well known that, if a vector space has a basis of $$n$$ vectors, it cannot have a basis with a different number of vectors. Thus, if you have $$n-1$$ linearly indipendent vectors, they cannot span $$V$$, otherwise $$V$$ would have a basis with $$n-1$$ elements, which is not possible.
A set of vectors spanning a space is a basis iff it is the minimum number of vectors needed to span the space. So if you reduce the number of vectors in your basis, it is no longer a basis for $$R^n$$ but will instead form a basis for $$R^{n-1}$$
You can prove this more rigorously by writing any $$x \in V$$ as the sum of vectors from your linearly independent list and showing that if you remove one vector from the list you can no longer do this, as you cannot write the nth element as some sum of the other elements.