I was able to prove that any two non-collinear vectors in $\Bbb R^2$ can form the basis for $\Bbb R^2$ but was unable to prove it for higher dimensions. My book didn't mention anything about dimensions higher than 2 so I don't even know whether the statement is even true.
If this is true, please provide a proof.
On
An alternative argument runs as follows. The parallelepiped formed by 3 non-coplanar vectors in $\mathbb{R}^3$ has non-zero volume. Therefore the determinant of the 3x3 matrix formed by the components of these vectors is non-zero. Therefore these 3 vectors are linearly independent. And therefore they are a basis for $\mathbb{R}^3$.
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A basis of a vector space is a set of vectors which are linearly independent and span the space. By Steinitz's exchange lemma, every basis of a vector space has the same length, which is what is called the dimension of the vector space. Thus, if you have a list of $n$ linearly independent vectors in a space for which the length of a basis is $n$, then those vectors form a basis and as a consequence span the space as well.
Yes, it is true. Three vectors do not span $\mathbb{R}^3$ if and only if they span a space whose dimension is at most $2$. In other words, if and only if the space that they span is contained in a plane. But this means that they are coplanar.