Suppose $T:V\to W$ is a linear map and $V$ is finite dimensional and $W$ is not necessarily finite dimensional. Is it still true that $Rank(T)=Rank(T^*)?$ Where $T^*$ is the dual map of $T.$
I know this would hold if $W$ is finite dimensional and so I thought about considering $T:V\to T(V)$ only rather than $W$ and then the 'range' would be finite dimensional. However I am not sure if this works and I am much interested in finding a better proof to this.
Many thanks!
It's true more generally that if $T : V \to W$ has finite rank then $\text{rank}(T) = \text{rank}(T^{\ast})$. To see this abstractly, $T$ has finite rank $n$ iff it factors as a composite
$$V \xrightarrow{f} R \xrightarrow{g} W$$
where $f$ is surjective, $g$ is injective, and $\dim R = n$. Taking duals gives that $T^{\ast}$ factors as a composite
$$W^{\ast} \xrightarrow{g^{\ast}} R^{\ast} \xrightarrow{f^{\ast}} V^{\ast}$$
and taking duals takes injective maps to surjective maps and vice versa (this is not entirely formal in the infinite-dimensional case because taking duals is not an equivalence, so it needs to be checked) and $\dim R = \dim R^{\ast}$ so $T^{\ast}$ also has rank $n$.
This is badly false once $T$ no longer has finite rank. For example the rank of the identity $I : V \to V$ is $\dim V$ but the rank of its dual $I^{\ast} : V^{\ast} \to V^{\ast}$ is $\dim V^{\ast}$ and this is generally larger than $\dim V$.