LA - Identity Maps and Injectivity

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Just one small remark: you say that "dim rangeT \leq dimW". This is not necessary, we have that dim rangeT = dim W because T maps onto W. But this is just a minor detail, your proof is fine and correct.In summary, we can prove that for a finite dimensional vector space V and a linear transformation T: V→W, T is injective if and only if there exists a linear transformation S: W→V such that ST is the identity map on V. This can be shown by extending a basis of V to a basis of W, and defining S to map the basis elements of W to their corresponding basis elements in V. This ensures that ST is the identity map on V,
  • #1
MushroomPirat
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Homework Statement


Suppose that [itex]W[/itex] is finite dimensional and [itex]T:V\rightarrow W[/itex]. Prove that [itex]T[/itex] is injective if and only if there exists [itex]S:W\rightarrow V[/itex] such that [itex]ST[/itex] is the identity map on [itex]V[/itex].


Homework Equations





The Attempt at a Solution



First, suppose that [itex]T[/itex] is injective and let [itex]Tu=Tv[/itex] for [itex]u,v\in V[/itex]. Clearly, [itex]Tu-Tv=0[/itex] and thus [itex]S(Tu-Tv)=0[/itex]. From this, we can see that [itex]STu=STv[/itex]. However, since [itex]T[/itex] is injective, then [itex]u=v[/itex]. Therefore, there exists an [itex]S[/itex] such that [itex]ST[/itex] is the identity map on [itex]V[/itex]. In the other direction, suppose [itex]ST[/itex] is the identity map on [itex]V[/itex], and let [itex]Tu=Tv[/itex]. From the previous argument, we can see that [itex]STu=STv[/itex], and thus [itex]u=v[/itex], so [itex]T[/itex] is injective.


I think the second part of my proof is right, going from identity map to injectivity, but I'm just not sure about my argument for the first half.

Thanks for your help!
 
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  • #2
MushroomPirat said:
First, suppose that [itex]T[/itex] is injective and let [itex]Tu=Tv[/itex] for [itex]u,v\in V[/itex]. Clearly, [itex]Tu-Tv=0[/itex] and thus [itex]S(Tu-Tv)=0[/itex]. From this, we can see that [itex]STu=STv[/itex]. However, since [itex]T[/itex] is injective, then [itex]u=v[/itex]. Therefore, there exists an [itex]S[/itex] such that [itex]ST[/itex] is the identity map on [itex]V[/itex].

I can't see how you concluded that. The only thing you can conclude is that STu=STu. The thing you must show is that STu=u!
In fact, you haven't even defined a suitable S! You must first define a candidate S, and only then can you show that ST is the identity. So, how would you define such an S?

In the other direction, suppose [itex]ST[/itex] is the identity map on [itex]V[/itex], and let [itex]Tu=Tv[/itex]. From the previous argument, we can see that [itex]STu=STv[/itex], and thus [itex]u=v[/itex], so [itex]T[/itex] is injective.

This is ok.
 
  • #3
Thanks for your help micromass. I thought about your advice and came up with a different proof:

Suppose T is injective and let [itex](v_1,...,v_n)[/itex] be a basis of V. Because this basis is linearly independent and T is injective, then [itex](Tv_1,...,Tv_n)[/itex] is also linearly independent (this was a result from a previous problem). Because T is injective, then dim V = dim rangeT [itex]\leq[/itex] dimW. Extend the previous list to a basis [itex](Tv_1,...,Tv_n,w_1,...,w_m)[/itex] of W. Now, define [itex]S: W\rightarrow V[/itex] such that [itex]STv_i=v_i[/itex] for all i=1,...,n, and [itex]Sw_j=0[/itex] for all j=1,...,m. Thus, if u is an arbitrary element of V, written as [itex]u=a_1v_1+...+a_nv_n[/itex] then [itex]STu=ST(a_1v_1+...+a_nv_n)=a_1v_1+...+a_nv_n[/itex] so ST must be the identity map on V.
 
  • #4
MushroomPirat said:
Thanks for your help micromass. I thought about your advice and came up with a different proof:

Suppose T is injective and let [itex](v_1,...,v_n)[/itex] be a basis of V. Because this basis is linearly independent and T is injective, then [itex](Tv_1,...,Tv_n)[/itex] is also linearly independent (this was a result from a previous problem). Because T is injective, then dim V = dim rangeT [itex]\leq[/itex] dimW. Extend the previous list to a basis [itex](Tv_1,...,Tv_n,w_1,...,w_m)[/itex] of W. Now, define [itex]S: W\rightarrow V[/itex] such that [itex]STv_i=v_i[/itex] for all i=1,...,n, and [itex]Sw_j=0[/itex] for all j=1,...,m. Thus, if u is an arbitrary element of V, written as [itex]u=a_1v_1+...+a_nv_n[/itex] then [itex]STu=ST(a_1v_1+...+a_nv_n)=a_1v_1+...+a_nv_n[/itex] so ST must be the identity map on V.

Now that looks good to me.
 

1. What is an identity map in LA?

An identity map in linear algebra refers to a function or transformation that maps a vector to itself. In other words, the input and output vectors are identical. It is denoted as I and represented by a diagonal matrix with ones on the main diagonal and zeros everywhere else.

2. How is an identity map useful in LA?

Identity maps are useful in linear algebra because they preserve properties of vector spaces and allow for easier calculations. For example, multiplying any vector by the identity map will result in the same vector, making it a useful tool in solving systems of equations and other linear transformations.

3. What is injectivity in LA?

Injectivity, also known as one-to-one mapping, is a property of functions or transformations in linear algebra where each output vector corresponds to a unique input vector. In other words, no two input vectors can result in the same output vector.

4. How do identity maps relate to injectivity in LA?

Identity maps are always injective, as each input vector is mapped to itself, and no two input vectors can result in the same output vector. However, not all injective maps are identity maps, as they can also map different input vectors to different output vectors.

5. What is the significance of injectivity in LA?

Injectivity plays a crucial role in determining the invertibility of linear transformations. If a transformation is injective, it means that it has a unique inverse, which is necessary for solving many problems in linear algebra. Additionally, injective maps preserve linear independence, making them useful in proving various theorems and solving equations.

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