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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.

mrtaurho
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Tutusaus
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  • Actually, (1) always holds, and it's basically the theorem that the row and the column rank of a matrix are the same. – Berci Apr 18 '20 at 15:09
  • Does this theorem you are telling me about have a specific name so I can look for it on the Internet? Thx. – Tutusaus Apr 18 '20 at 15:20
  • Maybe 'rank theorem'. I might be able to write you details later when I find more time for that.. The key is that the matrix representation of $f^*$ in the dual bases is the transpose of the matrix of $f$. – Berci Apr 18 '20 at 15:21

1 Answers1

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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.

Ben Grossmann
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