[Math] Rank-Nullity Theorem Proof.

Question

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?

Answer 1

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,

$$ \mathrm{dim}\ \mathrm{ker}\ \theta + \mathrm{dim}\ \mathrm{im}\ \theta = p + (n-p) = n = \mathrm{dim}\ V \ . $$