How to prove that if $\{v_1,v_2,..,v_n\}$ $n>1$ are solutions to $Ax=b$, $b \neq 0$ then $rank(A)=1$?

Question

Given $A_{m \times n}$, $\{v_1,v_2,..,v_n\}$ is a basis in $\mathbb{R^n}$, $n>1$ and $b \in \mathbb{R^m}$, $b \neq 0$, prove that if $\{v_1,v_2,..,v_n\}$ are solutions to $Ax=b$, then $rank(A)=1$?

Because $Ax=b$ has multiple solutions this means that $A$ is linearly dependent. That means that $m < n$. But it is given that $n>1$ so $m \le 1$.

Because $b \neq 0$ then $m \neq 0$ otherwise we'd get no solutions to $Ax=b\quad$ (1). Thus: $$m=1 \Rightarrow rank(A) \le min(m,n)\le m=1$$ If $rank(A)=0$ then $dim(\text{column space of A})=0$ but this cannot be because of (1) so $rank(A)=1$.

Is my proof exhaustive enough?

Answer 1

The fact that the columns of $A$ are linearly dependent doesn't imply $m<n$: the columns of the zero $m\times n$ matrix are linearly dependent for every $m$ and $n$.

Moreover, from $m<n$ and $n>1$ you cannot deduce that $m\le 1$.

Your attempt falls apart, sorry.

---

Since $v_n$ is a solution of $Ax=b$, you know that every solution of the linear system is of the form $v_n+v$, where $Av=0$. Now $v_i-v_n$ belongs to the null space of $A$ (because $v_n+(v_i-v_n)$ is a solution of $Ax=b$), for $i=1,2,\dots,n-1$.

This means that the subspace spanned by $\{v_1-v_n,\dots,v_{n-1}-v_n\}$ is contained in the null space of $A$. What's its dimension?

Can you finish?

Answer 2

Also, I think you should use the fact that solutions ${v_1,v_2,..,v_n}$ form a basis - this means that these vectors are linearly independent. So if $Ax=b$ has multiple independent solutions, then columns of A are linearly dependent.

Answer 3

As every element of $\Bbb R^n$ is a linear combination of the basis, $$A\left(\sum_{k=1}^n\lambda_k v_k\right) = \left(\sum_{k=1}^n\lambda_k\right)b,$$ i.e., $$\text{rank }A = \dim(\text{Im }A) = 1.$$