Proving Injectivity and the Identity Map for Finite Dimensional Linear Maps

  • Context: Graduate 
  • Thread starter Thread starter mind0nmath
  • Start date Start date
  • Tags Tags
    Identity Map
Click For Summary

Discussion Overview

The discussion revolves around proving that for a finite dimensional linear map \( T: V \to W \), \( T \) is injective if and only if there exists a linear transformation \( S: W \to V \) such that \( ST \) is the identity map on \( V \). Participants explore the connection between injectivity and the identity map, as well as the implications of dimensionality in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question the definition of injectivity and how it relates to defining a function from \( W \) to \( V \).
  • One participant provides a proof outline showing that if \( ST = I \), then \( T \) is injective, but raises a concern that the converse may only hold when the dimensions of \( V \) and \( W \) are equal.
  • Another participant argues that for \( T \) to be injective, it must be bijective, suggesting that injectivity alone does not guarantee the existence of an inverse \( S \).
  • Some participants discuss the implications of dimensionality, noting that \( \dim V \) must be greater than or equal to \( \dim U \) for injectivity, but equality is not necessary.
  • One participant suggests a method for defining \( S \) based on the preimage under \( T \) and addresses the potential for \( S \) to be zero for elements not in the image of \( T \).
  • Another participant clarifies that \( ST = I \) does not imply \( TS = I \), emphasizing the distinction between the mappings involved.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of dimensional equality for the injectivity and existence of an inverse. There is no consensus on whether the proposition is true or false, and the discussion remains unresolved regarding the implications of dimensionality.

Contextual Notes

Some participants note the importance of assumptions regarding the dimensions of the vector spaces involved, as well as the definitions of injectivity and linear transformations.

mind0nmath
Messages
19
Reaction score
0
The question is to prove for finite dimensional T: V to W,
T is injective iff there exists an S: W to V such that ST is the identity map on V.
I can't quite make the connection between injectivity and the identity map.
any suggestions?
thanks in advance.
 
Physics news on Phys.org
What's the definition of injectivity? How could you use that to define a function from W to V?
 
mind0nmath said:
The question is to prove for finite dimensional T: V to W,
T is injective iff there exists an S: W to V such that ST is the identity map on V.
I can't quite make the connection between injectivity and the identity map.
any suggestions?
thanks in advance.
Have you tried working with any specific examples to get ideas? Can you think of any examples of injective linear maps?

(I'm assuming V and W are supposed to be vector spaces, and T and S are supposed to be linear transformations)

(What does it mean for T to be "finite dimensional"?)
 
Suppose ST = I
To show T is injective:
Let T(x)=T(y) , f.s. x,y in V
=> S(T(x)) = S(T(y)) [permitted since S is well defined from W to V, and T is of course from V to W]
=> (ST)(x) = (ST)(y)
=> I(x) = I(y)
=> x = y
proving, T is injective.


I think the converse is true only when Dim V = Dim W


Now, suppose T is injective.
Let x be arbitrary in N(T)
T(x) = 0 (zero of W) = T(0)
=> x = 0
So, your N(T) = {0}
Now Dim N(T) + Dim R(T) = Dim V (Sylvester's law)
But N(T) = 0
=> Dim R(T) = Dim V = Dim W
Also R(T) is contained in W
Therefore, R(T) = W
=> T is surjective.

Now T is bijective => T is invertible, so there exists an S from W to V s.t ST = TS = I

Q.E.D.
 
You are assuming that U and V are finite dimensional vector spaces and T is a linear transformation from U to V?

Saying that there exist S such that ST= I means that T has an inverse. That is not possible unless T is, in fact, bijective: both injective and surjective. It is fairly easy to prove that if a linear transformation is injective, then it is also surjective, and vice-versa.
 
erm, where did my reply go ?
this is strange.
 
HallsofIvy said:
You are assuming that U and V are finite dimensional vector spaces and T is a linear transformation from U to V?

Saying that there exist S such that ST= I means that T has an inverse. That is not possible unless T is, in fact, bijective: both injective and surjective. It is fairly easy to prove that if a linear transformation is injective, then it is also surjective, and vice-versa.

I don't think this is true unless you assume U and V have the same dimension. For T to be injective, it must be the case that dim V \geq dim U, but equality doesn't have to hold. However, there does have to be an S such that ST is the identity on U (namely you just take Sx to be the preimage under T if x is in the image, and take Sx = 0 otherwise).

Of course, knowing me I'm probably overlooking something painfully obvious, so please be gentle if I'm wrong here.
 
No, you are completely right. I was thinking of U and V having the same dimension without even realizing it.
 
thanks for all the hints. To clear up: S and T are Linear Transformations. and W is finite dimensional (V and W are vector spaces though nothing mentioned about the dimension of V) . I remember from a logic/proof class that the way to prove if and only if statements is to go both ways meaning:
1) suppose T is injective. Show that there exists a linear transformation S: W -> V such that ST is the identity map on V.
2) Suppose there exists a linear transformation S: W -> V such that ST is the identity map on V. Prove T is injective.
I know that the question is hinting at an inverse of T (namely S), and I know that invertible functions are bijective. so is this proposition false? then what would be the counter example? of if true, how do I go from T is injective to an inverse S?
thanks again
 
  • #10
the entire proof is there !
scroll up to my reply.
 
  • #11
By the way, my example map is completely wrong. That's what happens when you answer algebra questions while doing complex analysis I suppose.
 
  • #12
what if V is not finite dimensional? how do you prove that T is injective implies that there exists S such that ST is an identity map?
 
  • #13
Hi,

The first half of sihag proof is right but the assumption that:
Dim V = Dim W,
is wrong.

The proof of the other way is:
If T:V->W is injective define
S(x) = y if T(y)=x or,
S(x) = 0 if for all y in V T(y) <> x
Hence
S(T(y)) = S(x) = x => ST is the identity for all element in V.

Just remember that ST = I does not mean TS = I. This is explained as
T: V->W and S:W->V,
ST: V -> V and TS: W -> W
they are different maps.
 

Similar threads

Graduate A problem in multilinear algebra
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Proof of Invertibility: Linear Map's Surjectivity and Injectivity Condition
  • · Replies 5 ·
Replies
5
Views
2K
MHB Injective linear mapping - image
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad DimKer(T) = 0 <--> T is injective
  • · Replies 6 ·
Replies
6
Views
5K
Undergrad Does this theorem need that Ker{F}=0?
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Explanation of a Line of a proof in Axler Linear Algebra Done Right 3r
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Can a Non-Linear Map be the Inverse of a Linear Map?
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Meaning of "passage to the quotient"?
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad ##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?
  • · Replies 31 ·
2
Replies
31
Views
3K
Graduate Linear Map T: Proving Existence of KerT=U if dimU≥dimV-dimW
  • · Replies 6 ·
Replies
6
Views
3K