Proof $(\text{im} A^\top)^\bot = \ker A$ We are in the euclidean vectorspace $\mathbb{R}^n$ with the euclidean scalar product.
Let $A \in \mathbb{R}^{m*n}, \text{im} A := \text{im} (\varphi_A)$ and $\ker A := \ker(\varphi_A)$.
Prove $$(\text{im} A^\top)^\bot = \ker A$$
I have no clue on where i should even start, so any help or hints are welcome.
 A: Hint: The main notational observation is that if the rows of $A$ are given by the vectors $\mathbf{a}_k$, then $\text{im}(A^\top) = \text{span}_{\Bbb R}(\mathbf{a}_k)$. [Why?]
For instance, the inclusion $(\text{im}( A^\top))^\bot \subseteq \ker A$ is equivalent to $$\mathbf{x}\cdot\mathbf{a}_k=0 \text{ for all }k \text{ and all } \mathbf{x}\in\ker(A).$$
I encourage you these two are indeed equivalent and verify the "main notational observation" above. Then the plan is to write a similar-looking statement for the other inclusion $(\text{im}( A^\top))^\bot \supseteq \ker A$, and prove both.
A: $\operatorname{ker}(A)=\{x\in\mathbb{R}^n: Ax=0\}$,
$\operatorname{ran}(A^\intercal)=\{A^\intercal y:y\in\mathbb{R}^m\}$,
$\big(\operatorname{ran}(A^\intercal)\big)^\perp=\{z\in\mathbb{R}^n: \langle z, A^\intercal y\rangle =0, \,\text{for all}\, y\in\mathbb{R}^m\}$.
where $\langle h,k\rangle:= k^\intercal h$ denotes the standard  inner product in $\mathbb{R}^n$.
Notice that $x\in\operatorname{ker}(A)$ iff
$$0=\langle Ax,y\rangle=y^\intercal Ax=(A^\intercal y)^\intercal x=\langle x, A^\intercal y\rangle$$
for all $y\in \mathbb{R}^m$. This equivalent to $x\in\operatorname{ker}(A)$ iff $x\in \big(\operatorname{ran}(A)\big)^\perp$.

This result is valid in any linear space with inner product, for example in Hilbert spaces.
