row space and kernel of a matrix A Given a real $m \times n$ matrix $A$ and vectors $x,z \in R^n$ how can I show that
$x \in \ker{A} \wedge x^Tz = 0   \quad \Rightarrow \quad \exists y \in R^m : z = A^Ty $ ?
I thought to start with
$Ax= 0$ and left multiply each side by a vector $ y \in R^m$ to obtain $y^TAx= 0$. The equation is now scalar and we can transpose both sides into $x^TA^Ty= 0$. Now I don't know how to go on.

You have to show that $z\in \text{Im}(A^T)$. Recall that $\ker(A)=\text{Im}(A^T)^{\perp}$. Hence $x\in \text{Im}(A^T)^{\perp}$. Since $x^Tz=0$, $\left\langle x,z\right\rangle =0$, so $x\perp z$. Hence $z\in (\text{Im}(A^T)^{\perp})^{\perp}=\text{Im}(A^T)$.
