Rank-Nullity Theorem Proof. Let $V$ and $W$ be linear vector spaces. Let $\theta$ be a linear map from $V$ to $W$. Why is $\dim(V) = \dim(\operatorname{Im}(\theta)) + \dim(\ker(\theta))$? I know that there is an isomorphism between $\operatorname{Im}(\theta)$ and $V/\ker(\theta)$, and that the cosets of $\ker(\theta)$ (members of $V/\ker(\theta))$ partition $V$. How can I deduce the relationship from this?
 A: Perhaps a better approach would be starting like this:


*

*Let $v_1, \dots , v_p$ be any basis of $\mathrm{ker}\ \theta$.

*Complete it till you have a basis for $V$: $v_1, \dots , v_p, v_{p+1}, \dots , v_n$.

*Then show that $\theta v_{p+1}, \dots , \theta v_n$ is a basis for $\mathrm{im}\ \theta$.


As a consequence,
$$
\mathrm{dim}\ \mathrm{ker}\ \theta + \mathrm{dim}\ \mathrm{im}\ \theta = p + (n-p) = n = \mathrm{dim}\ V \ .
$$
A: $\newcommand{im}{\operatorname{im}} \newcommand{\codim}{\operatorname{codim}}$
\begin{align*}\im(\theta)\cong_K V/\ker(\theta) & \Rightarrow \dim_K(\im(\theta))=\dim_K(V/\ker(\theta))  \\
& \Rightarrow \dim_K(\im(\theta)) = \codim_K(\ker(\theta)) \\
&  \Rightarrow \dim_K(\im(\theta))=\dim_K(V)-\dim_K(\ker(\theta)) \\\end{align*}
A: Using your idea, you need to show that in general, if $T\oplus S=V$, then $V/S\simeq T$.
Hint  Take a basis of $S$, say $\mathscr B$, and extend it to a basis of $V$, $\mathscr B\cup \mathscr B_0$. Show that $\mathscr B'=\{v+S:v\in\mathscr B_0\}$ is a basis of $V/S$.
Thus, when working with finite dimensional spaces, $\dim (V/S)=\dim V-\dim S$. Can you see how to conclude?
