Suppose $a_1,…,a_k$ and $b$ are vectors in $\mathbb{R}^n$, such that $b \ne 0$, and $a_1,…,a_k$ are different from each other.
Also, suppose the equation $x_1a_1+…+x_ka_k = b$ has an infinite amount of solutions.
Now, assume $k \ge n+1$.
Can we say that the set $\{a_1,…,a_k\}$ spans $\mathbb{R}^n$?
I think the answer is positive. The reason is as follows:
$k \ge n+1$ implies a matrix with more columns than rows. Given that $x_1a_1+…+x_ka_k = b$ has an infinite amount of solutions, we can build a matrix that comprises of this exact linear combination and be sure we'd get an infinite amount of solutions. In that matrix, $b$ represents any vector in $\mathbb{R}^n$, and since we'd get an infinite amount of solutions, it means that any vector in $\mathbb{R}^n$ is a linear combination of $\{a_1,…,a_k\}$, hence $\{a_1,…,a_k\}$ spans $\mathbb{R}^n$.
However, I'm not sure that I'm right, and in any case it doesn't seem like a well-structured proof to me. So, I'd like to know if I'm right, and if so, how can I build a structured proof out of this?
Best Answer
This is false. Here is a simple counterexample: let $n=3$, $k=4$, $a_1=(1,0,0)$, $a_2=(0,1,0)$, $a_3=(1,1,0)$, $a_4=(-1,-1,0)$ and $b=(1,0,0)$. Clearly $\{a_1,a_2,a_3,a_4\}$ does not span $\mathbb{R}^3$ and $\sum_{i=1}^4 x_ia_i=b$ has infinitely many solutions (in fact, $x=(1,0,\lambda,\lambda)$ is a solution for every $\lambda$).