Proving Linear Transformation as Composition of S & T

In summary, the conversation discusses the proof of the assertion that any linear transformation can be written as the composition of a surjective and injective transformation in one specific order. The proof provided by the speaker is valid for finite-dimensional spaces but may not hold for infinite-dimensional spaces. Additionally, the composition of two linear transformations may not always be commutative, meaning that the assertion is only valid for one specific order of composition (either TS or ST).
  • #1
arestes
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Homework Statement


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.


Homework Equations



standard theorems to define linear transformations (define the values on a basis). Ker(A)={v in E/ Av = 0}, etc

The Attempt at a Solution


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!
 

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  • #2


Hi there,

Thank you for your thorough explanation and for sharing your proof. It seems like you have a good understanding of the topic and have addressed all the necessary cases for finite-dimensional spaces. However, you are correct in your assumption that this proof may not hold for infinite-dimensional spaces. In order to prove this assertion for infinite-dimensional spaces, you would need to use the axiom of choice, which is not always a valid assumption in mathematics. Additionally, the proof may involve more complex and abstract concepts that may not be easily understood by those who are not well-versed in the topic of linear algebra.

To address your question about the validity of this statement in general, it is important to note that a linear transformation can only be written as the composition of an injective and surjective transformation in one specific order (either TS or ST), but not both. This is because the composition of two linear transformations may not always be commutative. In other words, if A = ST, then it may not necessarily be true that A = TS. Therefore, it is not possible to decompose a linear transformation into both ST and TS forms.

I hope this helps clarify the topic for you. Keep up the good work with your problem solving and exploration in linear algebra. Best of luck with your studies!
 

1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another in a way that preserves the properties of addition and scalar multiplication. In simpler terms, it is a function that transforms one set of points into another set of points in a straight line or plane.

2. How do you prove that a linear transformation is a composition of two other linear transformations?

To prove that a linear transformation is a composition of two other linear transformations, you must show that the output of the first transformation, when used as the input for the second transformation, produces the same result as the original transformation. In other words, if T and S are two linear transformations, and you want to prove that T can be represented as the composition of S and another transformation U, you must show that T(x) = U(S(x)).

3. What is the significance of proving a linear transformation as a composition of two other transformations?

Proving a linear transformation as a composition of two other transformations can help simplify complex transformations and make them easier to understand and analyze. It also allows for easier manipulation and calculation of the transformation.

4. What are some common techniques used to prove linear transformations as compositions?

Some common techniques used to prove linear transformations as compositions include using matrices and matrix multiplication, using properties of linear transformations (such as the preservation of addition and scalar multiplication), and using algebraic manipulation to show that the composition of two transformations produces the same result as the original transformation.

5. Can a linear transformation be represented as a composition of more than two other transformations?

Yes, a linear transformation can be represented as a composition of any number of other transformations. This is known as a chain of transformations, and it can be useful in breaking down complex transformations into simpler, more manageable ones.

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