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Suppose $T$ is a linear transformation from $\mathbb R^n$ to $\mathbb R^m$. Then there exist linear transformations $T_1:\mathbb R^n\to \mathbb R^r$ and $T_2:\mathbb R^r\to\mathbb R^m$ such that $T=T_2T_1$, where $r=\operatorname {Rank}(T)$.

I want a proof of this theorem.It is actually a version of rank factorisation theorem which states that Any $m\times n$ matrix can be expressed as the product of $2$ matrices each of rank $r$.This can be proved using matrix.But I am looking for a proof that uses its transformation point of view.I want a proof of this theorem using purely the concept of linear transformations.Can someone help me with this,I am yet not able to construct a proof from the concept of transformation only,although one can easily turn the matrix proof into transformation by translating the matrix into transformation,but that would not be an independent approach,and will seem artificial.

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  • $\begingroup$ @Rodrigo de Azevedo Can you help me to answer this question? $\endgroup$ Oct 25, 2019 at 14:34

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Since you're looking for an approach that avoids matrices, I will also avoid selecting bases.

With that in mind, one proof is as follows: let $\tilde T: \Bbb R^n \to \operatorname{im}(T)$ denote the map $T$ where we have restricted the codomain to the image $\operatorname{im}(T)$, so that $\tilde T$ is a surjective map.

Now, note that $\Bbb R^r \cong \operatorname{im}(T)$. So, let $\phi: \Bbb R^r \to \operatorname{im}(T)$ be a linear isomorphism. We note that $T$ can be broken down into a sequence of maps: in particular, $$ x \in \Bbb R^n \overset{\tilde T}{\mapsto} T(x) \in \operatorname{im}(T) \overset{\iota}\mapsto T(x) \in \Bbb R^m $$ where $\iota:\operatorname{im}(T) \to \Bbb R^m$ denotes the inclusion mapping. Thus, we have $$ T = \iota \circ \tilde T= \iota \circ \phi \circ \phi^{-1} \circ\tilde T. $$ With that, $T_1 = \phi^{-1} \circ \tilde T$ and $T_2 = \iota \circ \phi$ are maps satisfying the required conditions.

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  • $\begingroup$ Note that the choice of $\phi$ here is arbitrary, but ultimately necessary in order to produce matrices $T_1,T_2$. $\endgroup$ Oct 26, 2019 at 19:42

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