Image of linear map and the image of its dual Let $V$ and $W$ be vector spaces over a common ground-field $F$, not necessarily finite-dimensional.
Let $T : V \to W$ be a linear transformation.
Let $T^* : W^* \to V^*$ be its dual, given by $T^* : w^* \mapsto w^* \circ T$.
Can we compare $\dim \operatorname{im} T$ and $\dim \operatorname{im} T^*$?
 A: It is known (at least for $V=W$ but I'm sure it's true in general) that $T^*$ has finite rank if and only if $T$ has finite rank. In this case, they have same rank.
A: We can factor $T$ as
$$V \xrightarrow{\pi} V/\ker T \xrightarrow{\tilde{T}} \operatorname{im} T \xrightarrow{\iota} W$$
and correspondingly $T^{\ast}$ as
$$V^{\ast} \xleftarrow{\pi^{\ast}} (V/\ker T)^{\ast} \xleftarrow{\tilde{T}^{\ast}} (\operatorname{im} T)^{\ast} \xleftarrow{\iota^{\ast}} W^{\ast}.$$
Since $\pi$ is surjective, it follows that $\pi^{\ast}$ is injective, and since $\tilde{T}$ is an isomorphism, so is $\tilde{T}^{\ast}$. Using the axiom of choice (since we might not be able to talk about dimensions without it, there's no point trying to avoid it), the injectivity of $\iota$ implies the surjectivity of $\iota^{\ast}$. It follows that
$$\dim \operatorname{im} T^{\ast} = \dim\: (V/\ker T)^{\ast} = \dim\: (\operatorname{im} T)^{\ast} \geqslant \dim \operatorname{im} T\,,$$
with equality if and only if $T$ has finite rank.
