What is the Relationship Between the Image and Kernel of T and T^n?

  • Context: Graduate 
  • Thread starter Thread starter Chen
  • Start date Start date
  • Tags Tags
    Image Kernel
Click For Summary

Discussion Overview

The discussion revolves around the relationship between the image and kernel of a linear map \( T \) and its powers \( T^n \). Participants explore whether certain inclusions hold true for any natural number \( n \), specifically examining the implications for kernels and images in the context of linear algebra.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Chen proposes that \( \ker(T) \subseteq \ker(T^2) \) and \( \text{Im}(T) \supseteq \text{Im}(T^2) \) may hold for any natural \( n \) and questions how to prove this, suggesting induction as a method.
  • One participant confirms that \( \ker(T) \subseteq \ker(T^n) \) is easy to prove and provides an inductive argument for this inclusion.
  • Another participant clarifies the definitions of \( \ker(T) \) and \( \ker(T^2) \) and argues that induction may not be necessary if one accepts that \( T^n(0) = 0 \) for all \( n \).
  • A later reply discusses the relationship between images, stating that \( \text{Im}(T^2) = T(\text{Im}(T)) \) and provides a reasoning for the inclusion \( \text{Im}(T) \subseteq \text{Im}(T^2) \) and further inclusions for higher powers.

Areas of Agreement / Disagreement

Participants generally agree on the inclusion \( \ker(T) \subseteq \ker(T^n) \) and provide reasoning for it. However, there is no consensus on the necessity of induction for proving these inclusions or on the specifics of the image relationships, as different approaches are presented without resolution.

Contextual Notes

Some participants express uncertainty about the need for induction in proving the kernel inclusions, while others provide alternative reasoning. The discussion on images remains less formalized, with various interpretations of the relationships presented.

Chen
Messages
976
Reaction score
1
Hi,

What's the relationship between the image and kernel of T and the image and kernel of Tn? I think we saw in class something along the lines of:

[tex]Ker(T) \subseteq Ker(T^2)[/tex]
[tex]Im(T) \supseteq Im(T^2)[/tex]

My intuition is that this is also correct for any natural n, but is it true and if so how do you prove it, by induction?

Thanks,
Chen
 
Physics news on Phys.org
Assuming that T is a linear map from some vector space V to itself... [tex]\ker{T} \subseteq \ker{T^n}[/tex] (where n is any natural number) is easy to prove. It's obvious for n = 1.

Suppose it's true for n = k. If [itex]x \in \ker{T}[/itex], then [itex]x \in \ker{T^k}[/itex], by the induction hypothesis. But then [itex]T^{k+1}(x) = T(T^{k}(x)) = T(0) = 0[/itex] (by the linearity of T), so that [itex]x \in \ker{T^{k+1}}[/itex].

I think the thing about Im(T) could be done without induction.
 
Last edited:
Let me clear the cobwebs from out my skull...
Let T be a linear map from V to V'.

A_1=Ker(T)={v in V : Tv=0}.

A_2=Ker(T^2)={v in V : TTv=0}.

Note that T0=0 for all linear maps T. Here, the first 0 is in V and the second 0 is in V'.

Then A_1 is a subset of A_2 because:
v in A_1 implies
Tv=0 implies
TTv=T0=0
implies v in A_2.

I don't think you really need induction if you can get away with saying that (T^n)0=0. I suppose that technically you do need induction if it's not acceptable as being obvious: T0=0, so done when n=1. Then assuming (T^(n-1))0=0, we can apply T to both sides to get (T^n)0=T0=0. Done.

Then if you let A_n=Ker(T^n), go through the above proof to show that A_(n-1) is a subset of A_n (change 1 to n-1 and 2 to n).

Then you have the following result:
A_1 is a subset of A_2 is a subset of ... is a subset of A_n.

The image I'll work out if no one else does after I have a cigarrette...
 
Note that for a linear operator T on a vector space V, we have

[tex]Im(T^2) = T(T(V)) = T(Im(T))[/tex]

whereas [itex]Im(T) = T(V)[/itex]. Clearly, since [itex]Im(T) \subseteq V[/itex], [itex]T(Im(T)) \subseteq T(V)[/itex] follows immediately. For any n, we can compare [itex]Im(T^n)[/itex] with [itex]Im(T^{n+1})[/itex].

[tex]Im(T^{n+1}) = T^{n+1}(V) = T^n(T(V)) = T^n(Im(T))[/tex]

whereas [itex]Im(T^n) = T^n(V)[/itex]. Again, since [itex]Im(T) \subseteq V[/itex], it follows immediately that [itex]T^n(Im(T)) \subseteq T^n(V)[/itex], giving [itex]Im(T^{n+1}) \subseteq Im(T^n)[/itex]. This gives:

[tex]Im(T) \subseteq Im(T^2) \subseteq \dots \subseteq Im(T^n) \subseteq \dots[/tex]

which is what you wanted, I suppose.
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K