I have a question regarding dimensions of finite vectors spaces.
Let $f$ be a linear map between two vector spaces $f:E_{1}\longrightarrow E_{2}$ with dimensions $m$ and $n$ respectively in a field $\mathbb{F}$ and let $E_{1}^{*}$ and $E_{2}^{*}$ be their respective dual spaces. Is it true that $$\dim({\rm im}(f))\overset{(1)}=\dim({\rm im}(f^{*}))?$$
Where $f^{*}$ denotes the dual application, i.e. $f^{*}:E_{2}^{*}\longrightarrow E_{1}^{*}$.
I suspect that this affirmation is false and I have tried to prove it using a counter-example but unfortunately all the examples I have provided ended up supporting the statement (1).
I have a couple of other ideas to prove this statement but I don't know how to properly develop them. I suppose that by using the Corollary of the Isomorphism Theorem which states that $\dim(E_{1})\overset{\text{(2)}}{=}\dim(\ker(f))+\dim({\rm im})(f))$, where $E_{1}$ is a vector space, $f$ is a linear map $f:E_{1}\longrightarrow E_{2}$, $\ker(f)$ is the null space of $f$ and ${\rm im}(f)$ the image, one could end up proving (or not) this statement.
I have also come up with my own observations, which are:
\begin{align*} {\rm im}(f^{*})\subseteq E_{1}^{*} &\overset{\text{(2)}}{\implies} \dim({\rm im}(f^{*}))\leq \dim(E_{1}^{*})=\dim(E_{1})=m\\ {\rm im}(f)\subseteq E_{2} &\overset{\text{(2)}}{\implies} \dim({\rm im}(f))\leq \dim(E_{2})=n \end{align*}
And $n$ is not necessarily equal to $m$, so the statement is false.
Could we consider this a prove?
Anything will help.
Thanks in advanced.
The statement is true. One proof that this holds is as follows:
Let $r = \dim\operatorname{im}(f)$. Note that $f$ has a rank factorization. That is, there exists a surjective map $f_1:E_1 \to \Bbb F^r$ and an injective map $f_2:\Bbb F^r \to E_2$ such that $f = f_2 \circ f_1$. Note that $f^* = f_1^* \circ f_2^*$. Because $f_1$ is surjective, $f_1^*$ is injective. Because $f_2$ is injective, $f_2^*$ is surjective.
Because $f_2^*$ is surjective, we have $\operatorname{im}(f^*) = \operatorname{im}(f_1^*)$. Because $f_1^*$ is injective, $\dim\operatorname{im}(f_1^*) = r$. So, $\dim \operatorname{im}(f^*) = r$, which was the desired conclusion.