Solutions of a matrix equation and the orthocomplement of the kernel of the transpose of $A$ Let $A\in M_n(\mathbb R)$  and $b\in M_{n1}(\mathbb R)$ column vector. Prove $\exists x\in M_{n1}$ column vector solution to the equation $Ax=b\iff b=(\ker(A^{\tau}))^{\bot}$. 
So my idea was to first assume that there is such an $x$, so I want to show that $\langle b,y\rangle=0$ for all $y\in \ker(A^{\tau})$, but I dont know how to tackle it. 
This is not homework nor anything, so I dont mind if you spoil it. 
 A: There exists $x$ such that $Ax = b$  $\Longleftrightarrow$ 
$b$ is in the column space of $A$ $\Longleftrightarrow$
$b$ is in the row space of $A^{T}$ $\Longleftrightarrow$
Every element in the kernel of $A^{T}$ is orthogonal to $b$ $\Longleftrightarrow$
$b$ is in the orthogonal complement of the kernel of $A^{T}$

To see that the orthogonal complement of the kernel of $A^{T}$ is its row space, let $x \in \ker(A^{T})$. Then the column vector $A^{T}x$ has entries $r_{i} \cdot x$, the dot product of $x$ with the row vectors of $A^{T}$. If these are all zero, then $x$ is orthogonal to the rowspace of $A^{T}$. Conversely, if $x$ is orthogonal to the rowspace, all these $r_{i} \cdot x$ will be zero, so $x$ will be in the kernel of $A^{T}$.
A: Let $V \xrightarrow{A} W$ be a linear map between two inner product spaces $V$ and $W$.  Since they are inner product spaces there is the adjoint $V \xleftarrow{A^t} W$.  We have by rank nullity that $W=Ker(A^t) \oplus Im(A^t)^t=Ker(A^t)\oplus Im(A)$ where this is an orthogonal direct sum and I have used that $(A^t)^t=A$.  Hence $b \in Im(A)$ iff $b$ is orthogonal to $Ker(A^t)$.
