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!