Show that $dim\ rm{span} \{u_1,...,u_n\} = n-1 $ if and only if $dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$ Struggling with the following question:
Let $n \in \mathbb{N} $ with $n\geq 2$. Let $V$ be a vector space. Let $u_1,...,u_n \in V$. $\\$Show that $\dim\rm{span}  \{u_1,...,u_n\} = n-1 $ if and only if $\dim\{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0\} = 1$$\\$
My "attempt" for the forward direction:
Let $\left \{(a_1,...,a_n) \in \mathbb{R}^n \vert a_1u_1+...+a_nu_n=0 \right \} = W$
I believe we need to show that $\exists v \in W s.t. v \ne 0$ and $\rm{span}\{v \} = W $
I formed a basis of dimension $n-1$ from $\dim \rm{span} \left \{u_1,...,u_n \right \}$, call it $\beta$.
Now, am I supposed to show $\beta \in W$ and that $\beta$ spans $W$? This is where I'm lost. Any pointers or corrections are greatly appreciated.
Edit: Not allowed to use rank-nullity theorem.
 A: Consider a subset of $I\subset \{1, \ldots, n\}$ that $\{u_i \ | \ i \in I\}$ is linearly independent and $I$ maximal with this property.  May assume that $I= \{1,2,\ldots, k\}$. Now, for every $j>k$ the system $(u_1, \ldots, u_k, u_j)$ is linearly dependent, so we have
$$- \sum_{i=1}^k c_{j i} u_i + u_j = 0$$
Therefore, the $n-k$ vectors
$$\phi_j = (-c_{j1}, - c_{j2}, \ldots, - c_{jk}, \delta_{j, (k+1)}, \ldots, \delta_{j n}) $$
for $j = k+1, \ldots, n$,  are in the space $\Phi\colon= \{(a_1, \ldots, a_n) \ | \ \sum a_i u_i = 0\}$. Let us show that they form a base of this space. It should be clear that they are linearly independent, since the $(n-k) \times n$ matrix formed by them has as  an identity block $I_{n-k}$. Let us show linear independence:  Consider $\phi\colon = (a_1, \ldots, a_n)$, $\sum a_i u_i = 0$.  The vector
$$\phi-\sum_{j=k+1}^n a_j \phi_j $$
has the last $n-k$ components $0$ and gives a linear dependence of $u_i$, $i=1, n$. Hence it also gives a linear dependence of $u_1, \ldots, u_k$. But these $k$ vectors are linearly independent. It follows that the vector
$\phi-\sum_{j=k+1}^n a_j \phi_j $  has all its compoents $0$, that is
$$\phi = \sum_{j=k+1}^n a_j \phi_j$$.
A: Prove of $\Rightarrow$: From you hypothesis $\dim \rm{span}\{u_1,\ldots,u_n\} = n-1$. There is a linear independent subset that spans this very hyperplane, thus WLOG assume it is $\{u_1, \ldots, u_{n-1}\}$. Hence we have the unique decomposition of $u_n$ under this basis:$u_n = k_1u_1 + \ldots k_{n-1}u_{n-1}$.
Let $v=(k_1, \ldots, k_{n-1}, -1)$, then $v\ne 0$ and $v\in W$. Suppose $w=(l_1, \ldots, l_n)\in W$. If $l_n=0$, the by the definition of $W$ and the linear independence of $u_1, \ldots, u_{n-1}$, $l_1 = \ldots = l_{n-1} = 0$. Otherwise, $l_n \ne 0$, but $l_1 = \ldots = l_{n-1}$  is still uniquely determined by that unique decomposition in the previous paragraph: a multiple of $v$. This is enough to conclude $v$ spans $W$.
