# Thread: Decompose any linear transformation as a product of INJECTIVE and SURJECTIVE ones

1. ## Decompose any linear transformation as a product of INJECTIVE and SURJECTIVE ones

Hi!
I am solving problems from my linear algebra book and one of them asks me to prove that any linear transformation A: E ->F (between vector spaces E and F of any dimension, be it finite or not) can be written as the composition (product) of a surjective linear transformation S and an injective linear transformation T, that is: A = TS. ( A(x) =T ( S(x) ) )

I have done this, this is in fact just the same proof for the general fact that ANY FUNCTION can be expressed as the composition of a surjective and injective function in the same order as above .

HOWEVER, I was also asked to prove or disprove that any linear transformation can be also decomposed as the composition of an injective linear transformation T and a surjective linear transformation S, that is, A= ST . The reverse order.

Now, I will sketch a proof for the validity of this assertion, however, I think it relies on finite-dimensional reasoning and that's why I'm not quite sure if there's some subtle assumption I am overlooking there. I will in fact, assume that the vector spaces are finite-dimensional, but even with this assumption, I fail to see if there's any limitation to do the same for infinite-dimensional spaces (interpreting that dim(E)<dim(F) means that there is an injective linear transformation from E to F).

So, Let's take the arbitrary linear function A: E->F and assume that dim(E)<dim(F). Use R^n with n= dim(E) + dim(Ker(A)) + (dim(F) - dim(E) ) = dim(Ker(A)) + dim(F).
Now, consider a basis B of E. A must be defined just by considering its values on the elements of this basis. Consider the set A(B) = {Av, v in B}. Take a maximal linearly independent subset of A(B). (For finite-dimensional spaces this is just an inductive valid step, but I think I can do this in general by assuming the axiom of choice, in the form of Zorn's lemma, just like in the proof that every vector space has a basis).

so, I can modify the original basis in case by swapping the elements that are not in this maximal linearly independent subset by a suitable linear combination of them such that their images all go to 0 in F. So I can just assume that B is formed by elements whose images are L.I. and the others go to zero (they will generate the kernel).

Now, I define the a linear function T:E -> R^n by setting these elements to the first elements of the basis of R^n, which come a in number of dim(Ker(A)) (assuming finite dimensions) (because they are LI, I can do this without any contradictions).
the remaining elements of the basis of E are mapped to the next elements of the basis of R^n (they are a total of dim(E) - dim(Ker(A))).

In this way i define an injective (not necessarily surjective, because there are dim(F) - dim(E) elements of the basis of R^n that were not reached) linear transformation T.

Now, define the linear transformation S:R^n -> F by setting the first elements of R^n to zero so that A(x) = TS(x) for x being the elements of thebasis of E that originally
went to zero.

for the rest of elements of the basis of E that correspond to the next elements of the basis of R^n we map them to what they were originally. In this way A(x) = TS(x) for all elements of the basis of E, that is, A = TS.

But I still have room to cover all of F. I define the next dim(F)-dim(E) elements of the basis of R^n to be sent to the last corresponding elements of the basis of F. now we only need to cover the dim(Ker(A)) elements of F that were not reached because their corresponding ones were sent to zero. this can be covered by the last elements of the basis of R^n we have.

I know this is very long to follow but it's quite easy to see in a picture. I am attaching one.

Now, I know that this cumbersome proof should be ok for finite dimensional spaces but I see too many assumptions and uglyness for the general case for infinite dimensional spaces(with R^n replaced by a sufficiently big vector space). Does anyone know of a neat solution? Moreover, I did not address the cases where dim(E)>dim(F) (in the finite and infinite dimensional sense, even when one is finite and the other is not).

Is this false in general? (although i think the proof I showed above works for finite dimensional spaces). I was checking the wikipedia on surjection and injection and they mention the validity only for the decomposition A= TS, not the other way around (they say nothing, but if there were a symmetrical statement, they would mention it , right?)

any help regarding this would be very much appreciated.

thanks!

2. Oh wait, I see what you do in the proof now. You lift it first into a larger space for injection.

$A: E\to F$

Consider the product space $E \times F$ (no need to worry about finite dimensions).

$B: E\to E \times F$
where $B(e) = (e, A(e))$ injective.

$C: E \times F \to F$
where $C(e, f) = f$ surjective.

$A = CB$

Now this seems to be general enough to work for all functions and not just linear ones. Did I make a mistake?

3. Originally Posted by snowtea
Oh wait, I see what you do in the proof now. You lift it first into a larger space for injection.

$A: E\to F$

Consider the product space $E \times F$ (no need to worry about finite dimensions).

$B: E\to E \times F$
where $B(e) = (e, A(e))$ injective.

$C: E \times F \to F$
where $C(e, f) = f$ surjective.

$A = CB$

Now this seems to be general enough to work for all functions and not just linear ones. Did I make a mistake?
Thank you! This is completely right as far as I can see. But I suspect that it needs the Axiom of choice at some point. I'm not comfortable invoking it because I know it can be dispensed with sometimes when there is some rule to work with the infinities (not sure in which cases). Do you know how to make sure about its usage? In any case, this solves the problem. Thanks!

4. This argument is constructive - the injection and surjection are given explicitly. It looks like AC is not being used.