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.
 A: 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 $z\in (\text{Im}(A^T)^{\perp})^{\perp}=\text{Im}(A^T)$.
A: Consider a more general approach, in addition to the answer given.
Let $\mathcal{R}(A^T), \mathcal{N}(A)$ denote the row space and the nullspace of $A$, respectively.
We already know that the $2$ subspaces $\mathcal{N}(A)$ and $\mathcal{R}(A^T)$ are orthogonal. But it's not just that. $\mathcal{N}(A)$ and $\mathcal{R}(A^T)$ are orthogonal complements, according to the Fundamental Theorem of Linear Algebra, so we can say that: $$\big(\mathcal{N}(A)\big)^\perp = \mathcal{R}(A^T).$$
In short, the above states that whatever is orthogonal to the nullspace $\mathcal{N}(A)$, then it must belong to the row space.

$\newcommand{\N}{\mathcal{N}(A)}$
$\newcommand{\R}{\mathcal{R}(A^T)}$
The nullspace $\N$ includes all the vectors $x$ such that $Ax = 0$. Assume that there is a $z$, such that $z \perp \N$, but $z \notin \R$. Adding the vector $z$ as a last row of the matrix $A$ yields:
$$\underbrace{\begin{bmatrix} 
row \, 1 \\
row \, 2\\
\vdots\\
row \, m \\
coordinates \,of\, vector \, z
\end{bmatrix}}_{(m+1)\times n} \cdot \begin{bmatrix}
x_1\\
x_2 \\
\vdots\\
x_n
\end{bmatrix} =\begin{bmatrix}
0 \\
0 \\
\vdots\\
0
\end{bmatrix}
$$ 
The fundamental theorem of Linear Algebra states:

$$\dim \N + \dim \R = n.$$

We have to notice that although we modified the matrix $A$, $\N$ didn't change at all! $\N$ includes exactly the same vectors as before. Also, $n$ is the same as before.
This implies that $\dim\R$ must be the same as before, which means that $z \in \R$.
A: The point is that I want to prove that $\ker{A}$ and $im A^T$ are orthogonal complement.
It is easy to see that these spaces are orthogonal. Indeed if $z \in im A^T$ then we can write 
$z = A^T y$ and it follows $z \perp x, \ \forall x \in \ker A$, that is $z^T x = y^T A x = 0$.
Now from the fundamental theorem I can't understand why these two orthogonal spaces are complement of each other, that is my original question: why if a vector is orthogonal to  $\ker{A}$ it must lie on  $im A^T$ and vice versa? 
An example: consider three dimensional euclidean space (x,y,z). Consider the subspaces y=x and y=-x. They are orthogonal but not complement of each other because any vector on the z axis (x=y=0) is orthogonal to vectors on y=x but they don't lie on y=-x.
