Image and kernel of T^n

  • Thread starter Chen
  • Start date
  • #1
977
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
 

Answers and Replies

  • #2
695
0
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:
  • #3
1,569
2
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...
 
  • #4
AKG
Science Advisor
Homework Helper
2,565
4
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:

Related Threads on Image and kernel of T^n

Replies
12
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
7
Views
5K
Replies
5
Views
5K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
9
Views
6K
  • Last Post
Replies
3
Views
6K
Top