Prove that $\dim \operatorname{Nul}(A^t)=0$ and hence that $A$ has rank $m.$

Question

Let $A$ be an $m\times n$ matrix. Suppose that the equation $yA=0$ for a row vector $y$, a member of $\mathbb R^m$ has a unique solution. Prove that $\dim \operatorname{Nul}(A^t)=0$ and hence that $A$ has rank $m$.

I wasn't sure how to include the 'unique solution' information, as taking the transpose of the equation leads to $A^ty^t=0$, which is in the correct form to find the null space of $A^t$ more easily.

Answer 1

Hint: For any $A\in M_{m\times n}(\Bbb R)$ we have $\operatorname{Im}(A)\cap \ker(A^t)=\{\overline 0\}$. Can you proceed now?

Answer 2

Since $yA=0$ has a unique soloution and $y=0_{\mathbb{R}^{1\times m}}$ is a solution you have the unique solution. Now consider a solution $x_1\in \mathbb{R}^{m\times 1}$ of the equation $A^tx_1=0$ we will prove that $x_1$ has to be zero $$ A^tx_1=0_{\mathbb{R}^{n\times 1}} \iff (A^tx_1)^t=0_{\mathbb{R}^{n\times 1}}^t \iff x_1^tA=0_{\mathbb{R}^{1\times n}} $$ and $x_1^t\in \mathbb{R}^{1\times m}$. Therefore $x_1^t=0_{\mathbb{R}^{1\times m}}$ because of our hypothesis. So we get $x_1=0_{\mathbb{R}^{m\times 1}}$. So the only solution of the equation $A^tx=0$ is $x=0$ thus $Nul(A^t)=\{0\}\Rightarrow dim(Nul(A^t))=0$


Sidenote: If the eqution $Ax=0$ has a nonezero solution $x_1$ then $\lambda x_1\in span(x_1)$ is also a solution therefore you can find infinite solutions. In this problem the fact that the first equation has a unique solution implies the same for the transpose of the equation.

Answer 3

$Null (A^T) =\{x\in \mathbb R^m: A^Tx=0\}$
$yA=0$ has a unique solution and hence so does $A^Ty^T=0^T$ (Prove it!).
What is that unique solution? We observe that $y=(0,0,\dots,0)_{m\times 1}$ satisfies the above equation(homogeneous linear) . Therefore, $Null(A^T) =\{0\}$ and hence dimension of Null $(A^T) =0$