Rank-nullity theorem

Basis of the kernel. Let $(u_1,\ldots,u_k)$ be a basis of $\ker A$. Extend it to a basis $(u_1,\ldots,u_k,v_1,\ldots,v_{n-k})$ of $\mathbb{R}^n$.

Images of the $v_j$. The vectors $A v_1,\ldots,A v_{n-k}$ span $\operatorname{Im} A$: indeed, any vector in $\mathbb{R}^n$ is a combination of the $u_i$ and the $v_j$, and the $u_i$ are killed by $A$.

Independence. Suppose $\sum_j \alpha_j A v_j = 0$. Then $A(\sum_j \alpha_j v_j) = 0$, so $\sum_j \alpha_j v_j \in \ker A$. Since the $v_j$ are linearly independent from the $u_i$, this forces $\alpha_j = 0$ for all $j$.

Conclusion. The $A v_j$ form a basis of $\operatorname{Im} A$, so $\operatorname{rank}(A) = n - k = n - \dim(\ker A)$, i.e. $\dim(\ker A) + \operatorname{rank}(A) = n$.