Must a basis for an $n$-dimensional vector space have $n$ vectors? Does a basis for an $n$-dimensional vector space have to have $n$ vectors? For example, if I form a basis for $\mathbb{R}^n$, do I need at least $n$ vectors in my basis set?
In other words, can I form a basis for $\mathbb{R}^n$ by using only $n-1$, or less, vectors?
Note that, in this question, we only consider the whole vector space not creating a basis for a subspace.
 A: Yes, that is precisely the definition of the dimension. The number of vectors you need in a basis
A: Definition of dimension of a vector space is number of linearly independent vectors which will span the vector space. n-1 vectors may be linearly independent but they can not span the vector space.
A: If the columns of the $n\times n$ identity matrix were each a linear combination of a bunch of $n-1$ vectors $c_1$, $\ldots$, $c_{n-1}$,  then the determinant of the identity matrix, using the multi-linearity and skew-symmetry, would be $0$. 
Works for any commutative ring with $1$. 
A: There is this theorem about vector spaces which says that "If V has a basis with n elements then every set of vectors in V which has more than n elements is linearly dependent" . 
So Let W be a basis of V and S be a subset of it.
So W and S are both Linearly Independent.
We have to prove that S can never span V.
Keeping the above theorem in mind , we first on the contrary , assume that S also spans V . Then , by definition of basis , S is a basis of V. So by above theorem , any Set with size greater that size of Basis Set , is Linearly dependent , which contradicts that V is Linearly Independent . Hence , Must a basis for an n-dimensional vector space have n vectors.
Proved. 
A: The definition of basis of $\Bbb R^m$ is a set of vectors that are both linearly independent and spans $\Bbb R^m$. Assume that there are n vectors. Those n vectors in $\Bbb R^m$ form a matrix $[n_1 n_2 ...... n_n]$. If those vectors are linearly independent, that means that there must be n pivots in each column(because if there is a column without a pivot, there is a free variable, which leads to linear dependence). 
That shows n <= m.(if n>m, there will be columns without pivots and again linear dependence) If you are not convinced, write a $3 \times 2$ or $4 \times 2$ matrix and see if they are linearly dependent.
Analyzing the second definition--those vectors span $\Bbb R^m$. 
If that's the case, then the matrix will have a pivot in each row(if there is a row without a pivot, which has the form $[0,0,\cdots , 0]$, then the matrix can't span $\Bbb R^m$.) That gives us that $n \geq m$ .(again, try out some matrices that $n < $ m to see whether they span IR^m). 
combine those two conclusions, we have $n\leq m $ and $n \geq m$. So n must equal to m.
Therefore a basis has to have the form of $n \times n$.
