Is my proof correct: rank-nullity in a field $K$ Short motivation
I asked the question 
Dimension argument in the proof of Iversen's Theorem
where my ignorance to the subject of linear algebra was fully displayed. I want to make sure that I understand this correctly:
My rendering of the proof of the rank-nullity equation
Let $K$ be a field and let $A$ be an $m\times n$ matrix (height $m$ width $n$) defining a linear map $\varphi:K^n\rightarrow K^m$:
$$
A=
 \left[
  \begin{array}{ccc}
   a_{1,1} & \cdots  & a_{1, n}\\
   \vdots & \ddots & \vdots\\
   a_{m,1} & \cdots & a_{m,n}
  \end{array}
 \right]
$$
Let the column vectors of $A$ be denoted $a_1,a_2,...,a_n$. Then $\text{span}(a_1,...a_n)$ is a subspace of $K^m$ of dimension $\text{rank}(A)\leq \min(m,n)$. Clearly the nullspace of $A$ is a subspace of $K^n$ for if $x,y$ maps to zero, then $A(sx+ty)=0s+0t=0$ by distributive properties.
Now the pre-image $\varphi^{-1}(\varphi(K^n))$ must have dimension at least $\text{rank}(A)$ because the map $\varphi$ clearly is well defined. And if $x,y\in K^n$ maps to the same element in $K^m$ we get $A(x-y)=0$ so that $(x-y)$ belongs to the nullspace of $A$. This shows that any two different elements of the orthogonal complement of the nullspace of $A$ maps to different elements of $K^m$. Hence the dimension of the orthogonal complement of the nullspace equals the rank of $A$ so that
$$
\text{rank}(A)+\text{nullity}(A)=n
$$
or put differently
$$
\dim(\text{Im}(\varphi))+\dim(\text{Ker}(\varphi))=n
$$
Does this rendering make sense someone more skilled in the subject of Linear Algebra than I am?
 A: The main problem with your proof is that there is no such thing as the orthogonal complement in general vector spaces. This is particularly annoying if $K$ is a finite field (which you seem interested in), since there is no such thing as positive definite inner products in nonzero characteristic.
However, you can make a proper proof by choosing a complement to the kernel (orthogonality plays no role). Here is one way to do it. Choose a basis of $\def\Im{\operatorname{Im}}\Im(\varphi)$, and for every $w_i$ in that basis a vector $v_i$ in its pre-image: $\varphi(v_i)=w_i$. Let the subspace $C\subseteq V$ be the span of the $v_i$ (this will be our complement of $\ker(\varphi)$). Define $g:\Im(\varphi)\to C$ as the linear map such that $g(w_i)=v_i$ for all$~i$ (such a requirement on the basis of the $w_i$ extends uniquely by linearity). By the definition of$~C$, this linear map is surjective.
From $\varphi(v_i)=w_i$ we get that $\varphi|_C\circ g$ is the identity on $\Im(\varphi)$, so $g$ is injective, and gives a bijective linear map $\Im(\varphi)\to C$ whose inverse is the restriction $\varphi|_C$. Bijectivity gives $\dim C=\dim\Im(\varphi)$, and injectivity of the inverse $\varphi|_C$ gives that $\def\Ker{\operatorname{Ker}}C\cap\Ker(\varphi)=\{0\}$. Also $C+\Ker(\varphi)=K^n$ since for all $x\in K^n$ one has $x=g(\varphi(x))+(x-g(\varphi(x)))$ and the parenthesised subexpression is in $\Ker(\varphi)$. Then $K^n=C\oplus\Ker(\varphi)$, implying the rank-nullity theorem by the formula for the dimension of a direct sum.
A: Your argument makes sense to me (though it's sounds like a sketch of a proof). To make it more clear, I would recommend you to rewrite it in the following order.


*

*Confirm that $\ker\varphi$ is a subspace of $K^n$.

*Pick linearly independent elements that spans $\ker\varphi$, say $v_1, \cdots, v_k$.

*Extend those to $v_1, \cdots, v_k, w_1, \cdots, w_r$ so that these become a basis of $K^n$.

*Confirm that $K^n = \mathrm{span}\{v_1, \cdots, v_k\} \oplus \mathrm{span}\{w_1, \cdots, w_r\}$.

*Confirm that $\mathrm{im}\,\varphi$ is a subspace of $K^m$.

*Confirm that $\varphi(w_1), \cdots, \varphi(w_r)$ are a basis of $\mathrm{im}\,\varphi$.

*Conclude that $\dim(\mathrm{im}\,\varphi) + \dim(\ker\varphi) = n$.

