# Can you always write transformations $T=T_2 \circ T_1$ for some linear maps $T_1:V\to W$, $T_2:W\to V$?

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.

• 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.

• 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. Nov 17, 2020 at 16:00
• @Phi You're not missing anything; I got that mixed up when I wrote this answer. I'll try to fix that now Nov 17, 2020 at 16:03
• Here I have found one such $:$ math.stackexchange.com/a/3729053/778190 Please have a look at it. Thanks again. Nov 17, 2020 at 16:07
• @Phibetakappa Great! That works for me. Yes, that construction works 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 $$S\circ\pi=T$$.