Left ideals of matrix ring over a field The claim is made here that for $k$ a field

The left ideals of $M_n(k)$ are all of the form  $$\{A \in M_n(k) \mid
 \operatorname{ker}A > S \}, \rlap{ \qquad \text{for some subspace
 $S$.}}$$

I was trying to think through that claim. I understand why all the left ideals of $M_n(k)$ are of the form
$$\{A \in M_n(k) \mid \operatorname{Rowspace}(A) < S \}, \rlap{ \qquad \text{for some subspace $S$.}}$$
In the presence of an inner product, we have that $\operatorname{ker}A^T = \operatorname{Range}(A)^{\bot}$, and therefore we get the characterization we want. But what about if there isn't an inner product?
Does the same hold for $M_n(\Delta)$, where $\Delta$ is a division ring?
 A: Using the theory of nondegenerate symmetric bilinear forms, it is possible to show that $\operatorname{ker}A^T = \operatorname{Range}(A)^{\bot}$ without the need for existence of an inner product (when we are discussing matrices over a field). The symbol $\bot$ would then represent an orthogonal complement with respect to the relevant nondegenerate symmetric bilinear form. Here is a copy from some work of mine showing that the row space of a matrix is the right orthogonal complement of the null space with respect to what I would regard as the most natural nondegenerate symmetric bilinear form:

Consider the function $f: \mathscr{F}^n \times \mathscr{F}^n
> \rightarrow \mathscr{F}$, defined by $f(v,w)=v^Tw$. It is easy to see
  that:
  
  
*
  
*$f_{w_0}:\mathscr{F}^n \rightarrow \mathscr{F}$ defined by $f_{w_0}(v)=f(v,w_0)$ for any given $w_0 \in \mathscr{F}^n$ is a
  linear functional and therefore belongs to the dual space of
  $\mathscr{F}^n$. 
  
*Similarly $f_{v_0}:\mathscr{F}^n \rightarrow \mathscr{F}$ defined by $f_{v_0}(w)=f(v_0,w)$ for any given $v_0 \in \mathscr{F}^n$ is a
  linear functional and therefore belongs to the dual space of
  $\mathscr{F}^n$. 
  
  
  By the properties above, $f$ is a bilinear form, and furthermore it is
  a symmetric bilinear form since $f(v,w)=f(w,v)$ for any $v,w \in
 \mathscr{F}^n$. A symmetric bilinear form $g:\mathcal{V} \times
 \mathcal{V} \rightarrow \mathscr{F}$ is non-degenerate if and only if,
  for every nonzero $v \in \mathcal{V}$ there exists $w \in \mathcal{V}$
  so that $g(v,w) \neq 0$ [p.455]{golan}. Let
  $v=(v_1,v_2,\ldots,v_n)^T \in \mathscr{F}^n$ be nonzero. Suppose $i
 \in \{1,2,\ldots,n\}$ is the least integer so that $v_i$ is nonzero.
  Let $w=(0,\ldots,0,v_i,0,\ldots,0)^T \in \mathscr{F}^n$ (with $v_i$ in
  entry $i$ of $w$). Then $$f(v,w)=v^Tw=v_i^2 \neq 0,$$ since a field
  contains no divisors of zero. This proves that $f$ is non-degenerate
  for any field $\mathscr{F}$.
The right $f$-orthogonal complement [p.458]{golan} of row$(G)$ is
  \begin{eqnarray} \nonumber \text{row}(G)^{\perp_f}&=&\{w \in
 \mathscr{F}^n:f(v,w)=0 \text{ for all } v \in \text{row}(G) \} \\
 \nonumber &=& \{w \in \mathscr{F}^n:Gw=0 \} \\  &=& \text{N}(G).
  \end{eqnarray}

The theory can be studied in detail on page 455-458 of this textbook, which is the source referenced above. As you can see $f(v,w)$ is nondegenerate due to the fact that a field has no divisors of zero. 
For a division ring the symmetric property would fail in general, but I don't actually think this is absolutely necessary: You can see in the proof above I mention the right $f$-orthogonal complement, and the last part would still hold in the absence of commutativity.
