Let $S = {v_1, · · · , v_k}$ be a set of $k$ vectors in $\Bbb{R}^n$. Explain why $S$ cannot be a basis for $\Bbb{R}^n$ if $k > n$. Also, explain why $S$ cannot be a basis for $\Bbb{R}^n$ if $k < n$. (Thus, now we know that a basis of $\Bbb{R}^n$ must have $n$ vectors.)
I think the question implies it needs to have no free variables and pivot positions in every row, but I'm not sure. I understand there needs to be the same amount of vectors as the dimension space, but why I'm not exactly sure.
On
If R^n has a linearly independent subset of cardinality n+1 (a potential basis), it has a subspace isomorphic to R^(n+1) (the span of those n+1 vectors) we will call V^(n+1). But projecting V^(n+1) into R^(n+1) gives an n+1 element linearly independent set in R^(n+1) that is not a basis. Extending this to a basis yields a subspace isomorphic to R^(n+2). Iterating shows that R^n has subspaces isomorphic to each R^k. This can be easily shown to be false.
If $k>n$, then the basis is not linearly independent, that is, there exists an $x$ such that $v_x$ can be expressed as a linear combination of other elements of $S$.
If $k<n$, then the basis does not span the space. That is, there exists a $w \in \mathbb{R}^n$ that cannot be expressed as linear combination of elements of $S$.