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Let $T,U,V,W$ be vector spaces over $\mathbb F$ and let $\alpha:T \to U$, $\beta : V \to W$ be fixed linear maps. If the spaces are finite-dimensional and $\alpha$ and $\beta$ have rank $r$ and $s$ respectively, find the rank of the linear map $\Phi : \mathcal L(U,V) \to \mathcal L(T,W)$ which sends $\theta$ to $\beta \circ \theta \circ \alpha$.

I have come up with $\dim U \dim V - r (\dim V - s)$ by noting that $\theta$ must map the image of $\alpha$ to the kernel of $\beta$ to be the zero mapping, then using rank-nullity theorem, but I'm not sure if that's the answer that they are looking for. Is it possible to express it solely in terms of $r$ and $s$?

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$\newcommand{\range}{\operatorname{range}}\newcommand{\rank}{\operatorname{rank}} \newcommand{\card}{\operatorname{card}}$ Suppose the $u_1, \cdots, u_r$ is a basis for $\range \alpha $ and $w_1,\cdots, w_s$ is a basis for $\range \beta$. $\rank \Phi = \card \alpha^{-1}(\{u_1,\cdots, u_r\})\times \card \{w_1\cdots,w_s\}= (\dim T-\dim \ker\alpha)\times \dim \range\beta= \dim\range \alpha\times\dim\range\beta=rs. $

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According to https://en.wikipedia.org/wiki/Kronecker_product

$\phi=\beta\otimes \alpha^T$, if we stack the matrices row by row.

Then $rank(\phi)=rank(\beta)rank(\alpha^T)=rank(\beta)rank(\alpha)=rs$.

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  • $\begingroup$ Ah, ok. So if my formula is correct then I just need to show $\dim U =r$? But surely this is not true in general, as $\alpha$ need not be surjective. $\endgroup$
    – user85798
    Oct 25 '17 at 12:29

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