Prove that n-1 vectors cannot span a n dimensional vector space I want to prove that n-1 vectors in a n-dimensional space (say $\mathbb{R}^n$) cannot span the n-dimensional space. How can I do this?
One way is to find a vector that cannot be obtained as a linear combinations of the $n-1$ vectors. But how can I find this vector? (I cannot use concept of fundamental subspaces)
Another way is to assume that $n-1$ vectors span $\mathbb{R}^n$ and arrive at a contradiction. What contradiction can I find?
I'm allowed to use only definitions of vector space, subspace, linear independence and span of a set of vectors. I'm not allowed to use the concepts of basis, fundamental subspaces, etc. Is it even possible to prove this?
 A: Suppose for contradiction:
the following have the same span, with standard basis vectors $\mathbf e_k$
$\mathbf {B}  :=\bigg[\begin{array}{c|c|c|c|c} \mathbf e_1 & \mathbf e_2 &\cdots & \mathbf e_{n-1} & \mathbf e_{n}\end{array}\bigg]$ and   $\mathbf {B}' :=\bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2 &\cdots & \mathbf v_{n-1}\end{array}\bigg]$
Since the span is the same, each column in $\mathbf B$ may be written as a linear combination of columns of $\mathbf B'$.  That is,
$\mathbf B = \mathbf B' A$
where $A$ is short ($n-1$ rows) and fat ($n$ columns) but
$\mathbf B\mathbf y = \mathbf 0\implies \mathbf y = \mathbf 0$ by linear independence of standard basis vectors, yet
$A\mathbf y =\mathbf 0$ for some $\mathbf y \neq \mathbf 0$ which is a contradiction.
How do you know there is a non-zero $\mathbf y$ in the kernel of $A$? Suppose its first $n-1$ columns are linearly dependent-- then use the definition of linear dependence to get $\mathbf y$.  Alternatively suppose its first $n-1$ columns are linearly independent, then use row reduction or Cramer's Rule on that $n-1 \times n-1$ matrix to solve $A_{n-1\times n-1}\mathbf x = -\mathbf a_n$ and set $y_i:= x_i$ for $1\leq i \leq n-1$ and set $y_n:=1$.
A: Suppose for contradiction that $$S = \{\vec{v_1}, \vec{v_2}, ..., \vec{v_{n-1}}\}$$ spans $\mathbb{R}^n$.
Note that if $\vec{x}$ is a linear combination of the first $n-2$ vectors in $S$, then $$\text{span}\{\vec{v_1}, \vec{v_2}, ... , \vec{v_{n-1}}\} = \text{span}\{\vec{v_1}, \vec{v_2}, ... , \vec{v_{n-1}}+\vec{x}\}$$ To see this, call the set on the right-hand-side $S'.$
Any vector in the span of $S'$ is certainly in the span of $S$. Given any $c_1\vec{v_1}+...+c_{n-1}(\vec{v_{n-1}}+\vec{x})$, write $\vec{x}$ in terms of the other $\vec{v_i}$'s, expand, and simplify.
Moreover, any vector in the span of $S$ is in the span of $S'$. Any vector in the span of $S$ can be written as $$c_1\vec{v_1}+...+c_{n-1}\vec{v_{n-1}}=c_1\vec{v_1}+...+c_{n-1}(\vec{v_{n-1}}+\vec{x}-\vec{x})
= c_1\vec{v_1}+...+c_{n-1}(\vec{v_{n-1}}+\vec{x})-c_{n-1}\vec{x}$$ Since $\vec{x}$ is only a combination of the first $n-2$ vectors in $S$, the last term in the sum above merely changes the coefficients of $\vec{v_1},...,\vec{v_{n-2}}$.
Given a set of vectors, we have shown that we can add to one vector a combination of the others without changing the set's span. We can show that, under our initial assumption, $$\text{span}(S) = \text{span}\{\vec{e_1}, \vec{e_2}, ... , \vec{e_{n-1}}\}$$ This will contradict the fact that $S$ spans $\mathbb{R}^n$.
Consider the following procedure, which starts with $\vec{e_1}$. If $\vec{e_1}$ is in $S$, then move to $\vec{e_2}$. If $\vec{e_1}$ is not a vector in $S$, then there is exactly one set of constants satisfying $c_1\vec{v_1}+...+c_{n-1}\vec{v_{n-1}}=\vec{e_1}$. Choose $i$ such that $c_i\neq{0}$. Replace $\vec{v_i}$ with $\vec{e_1}$. If we call the new set $S_1$, then $\text{span}(S_1)=\text{span}(S)$. Move to $\vec{e_2}$. If $\vec{e_2}$ is in $S_1$, move on. Otherwise, there is exactly one set of constants satisfying $c_1\vec{v_1}+...+c_i\vec{e_1}+..+c_n\vec{v_n}=\vec{e_2}$. Choose $j$ such that $c_j\neq{0}$ and $j\neq{i}$. Such a $j$ must exist because $\vec{e_2}$ cannot be a multiple of $\vec{e_1}$. Replace $\vec{v_j}$ with $\vec{e_2}$. If our new set is $S_2$, then $\text{span}(S_2) = \text{span}(S_1) = \text{span}(S)$. Continue changing the elements of $S$ in this way, while not changing the span. Eventually, the set is changed to $S_{n-1} = \{\vec{e_1}, \vec{e_2},...,\vec{e_{n-1}}\}$.
