# Can any linear operator be written as composition of linear maps?

Let $$V$$ be a finite dimensional vector space over $$\mathbb{R}$$ and $$T:V\to V$$ be a linear map. Can you always write $$T= T_2\circ T_1$$ 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

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

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

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

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

One or more than one option are correct.

Do I have to show that any n×n matrix $$A$$ can be written as products of some n×m matrix $$B$$ and some m×n matrix $$C$$ such that one or more than one of the above options holds?

I have no clue then how to proceed.

Any help would be great. Thanks.

Edit: I think(after reading the question for almost 1 hour) that I have to choose different $$W$$ in each case otherwise the problem is absurd.

So , the first case and the 2nd case are false if I consider the zero transformation , whatever $$W$$ I choose, it doesn't matter .

I am working on case 3 and 4 , so if anyone comes up with anything please let me know.

• Is $W$ given? – Bernard Jul 15 '19 at 10:57
• What are your thoughts? Are you allowed to pick a different $W$ each time? – Joppy Jul 15 '19 at 10:59
• $W$ is exactly what I write above. and I wrote my thought above too. ..and but I don't know whether my thought is right or wrong. – suchanda adhikari Jul 15 '19 at 11:16

Hints:

1. The composition of two injective (resp. surjective) maps is injective (resp. surjective).

If $$W$$ is not given:

1. For case 3, think of $$\operatorname{Im}T$$.
2. For the last case, let $$E$$ any vector space. Look at $$W=V\oplus E$$.
• @ Bernard sir, I didn't understand your hints first, but now after spending a quality time with this problem I finally got your hints ...for the last option I realised that dim of $E$ has to be greater than or equal to dim of $V$. I want to know your opinion about that. – suchanda adhikari Jul 15 '19 at 18:30
• I don't think so, as there is always a canonical injection from each factor into their direct sum, and canonical projections from the direct sum onto each factor. This is independent of the dimensions.. I'm glad you've understood my hints, as they're intended to make the O.P. think about the problems, not to solve for him/her. – Bernard Jul 15 '19 at 18:37
• sir, sorry for disturbing you again, I understood what you said above but I didn't understand how it's gonna work in any arbitrary linear map $T$ . For example, suppose $T$ is the zero transformation then for the last case I have to find a suitable $W$ and a one one linear map $T_1$ and an onto linear map $T_2$ such that $T_2\circ T_1=T$ . I chose $W=E\oplus V$ where $E$ contains range of $T_1$ and then define $T_2$(x+v)=v for all x in E . Thus I can write $T_2\circ T_1=T$. So I don't understand how can I take any vector space $E$ . – suchanda adhikari Jul 16 '19 at 14:44
• Please sir if you please explain a little more. – suchanda adhikari Jul 16 '19 at 14:46
• You're right, I didn't explain enough this point. I only meant, as a hint, that, given a direct sum, we have canonical injections, and canonical projections, that you can choose $E$, so that $W=V\oplus E$ and linear maps $T_1, T_2$ satisfying the required conditions. For your example ($T=0$), just take $E=V$, $T_1 =$ the injection of the first factor, and $T_2=$ the second projection. – Bernard Jul 16 '19 at 14:58