Let $V$ be a finite-dimensional vector space over $\mathbb R$ and $T: V\to V$ be a linear map. Can you always write transformations $T=T_2 \circ T_1$ for some linear maps $ T_1:V\to W $, $ T_2:W\to V $, where $W$ is some finite-dimensional vector space and such that

A. both $T_1$ and $T_2$ are onto

B. both $T_1$ and $T_2$ are one to one

C. $T_1$ is onto, $T_2$ is one to one

D. $T_1$ is one to one , $T_2$ is onto

My Try Let $T=O,$ So, Range($T$)=$\{0\}$ and Ker($T$)=$V$. $O=O\circ T=O\circ O.$ I am getting $T_1$ and $T_2$ neither one-one nor onto. Not able to judge the options. Please help me.


2 Answers 2

  • Choice A: If $T_1,T_2$ are onto, then $T_1 \circ T_2$ will also be onto. So, since $T$ might not be onto, we cannot guarantee that there exist such onto maps $T_1,T_2$. For instance, if $T(x,y) = (x,0)$, then there are no onto maps $T_1,T_2$ such that $T_1 \circ T_2 = T$.

  • Choice B: Likewise, if $T_1,T_2$ are both one to one, then $T_1\circ T_2$ will also be one to one.

  • Choice C: Yes, this is always possible. Such maps $T_1,T_2$ form a rank factorization of $T$. The other answer explains the construction $$ V \overset{T}\to W = V \overset{\pi}{\to} V/\ker(T) \overset{S}\to W. $$ Another such decomposition is $T = \iota \circ \tilde T$ where $\tilde T: V \to \operatorname{im}(T)$ is defined by $\tilde T(v) = T(v)$ (but is onto because of the change in domain), and $\iota:\operatorname{im}(T) \to W$ is the inclusion map. That is, $$ V \overset T\to W = V \overset {\tilde T} \to \operatorname{im}(T) \overset \iota \to W. $$

  • Choice D: see this post.

  • $\begingroup$ I don't understand Choice D i.e. $T = \iota \circ \overline {T}.$ So according to the given question we have $T_2 = \iota$ and $T_1 = \overline {T}.$ Here $T_2$ is injective and $T_1$ is surjective which is Choice C. But Choice D states the other way round. Am I missing something? Thanks. $\endgroup$ Nov 17, 2020 at 16:00
  • $\begingroup$ @Phi You're not missing anything; I got that mixed up when I wrote this answer. I'll try to fix that now $\endgroup$ Nov 17, 2020 at 16:03
  • $\begingroup$ Here I have found one such $:$ math.stackexchange.com/a/3729053/778190 Please have a look at it. Thanks again. $\endgroup$ Nov 17, 2020 at 16:07
  • 1
    $\begingroup$ @Phibetakappa Great! That works for me. Yes, that construction works $\endgroup$ Nov 17, 2020 at 16:10

$V/\ker{T}\cong\text{Im}(T)$, now let $\pi:V\rightarrow V/\ker{T}$ be the canonical map, $\pi:v\rightarrow v+\ker{T}$ and $S:V/\ker{T}\rightarrow V$ by $S:v+\ker{T}\rightarrow T(v)$, then $S$ is one-to-one and $\pi$ is onto such that $S\circ\pi=T$.


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