Linear Algebra: Prove the orthogonal complement of the row space of A is {0} implies Ax = 0 has only the trivial solution

Question

Prove the orthogonal complement of the row space of $A $is ${0}$ implies $Ax = 0$ has only the trivial solution

Answer 1

Let us consider that $A\in M_{n}(\textbf{F})$ and denote by $R\leq \textbf{F}^{n}$ the row space of $A$. Once $R^{\perp} = \{0\}$, it results that \begin{align*} \textbf{F}^{n} = R\oplus R^{\perp} \Rightarrow \dim\textbf{F}^{n} = \dim R + \dim R^{\perp} = \dim R + 0 = \dim R \end{align*}

whence we conclude that $\text{rank}(A) = n$, and consequently that $A$ is invertible. Finally, one has that \begin{align*} Ax = 0 \Leftrightarrow A^{-1}(Ax) = A^{-1}0 \Leftrightarrow (AA^{-1})x = 0 \Leftrightarrow x = 0 \end{align*}

and we are done.

Hopefully this helps.