Is my proof correct: rank-nullity in a field $K$

Solution 1:

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.

Solution 2:

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.

1. Confirm that $\ker\varphi$ is a subspace of $K^n$.
2. Pick linearly independent elements that spans $\ker\varphi$, say $v_1, \cdots, v_k$.
3. Extend those to $v_1, \cdots, v_k, w_1, \cdots, w_r$ so that these become a basis of $K^n$.
4. Confirm that $K^n = \mathrm{span}\{v_1, \cdots, v_k\} \oplus \mathrm{span}\{w_1, \cdots, w_r\}$.
5. Confirm that $\mathrm{im}\,\varphi$ is a subspace of $K^m$.
6. Confirm that $\varphi(w_1), \cdots, \varphi(w_r)$ are a basis of $\mathrm{im}\,\varphi$.
7. Conclude that $\dim(\mathrm{im}\,\varphi) + \dim(\ker\varphi) = n$.