## Image and kernel of T^n

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:

$$Ker(T) \subseteq Ker(T^2)$$
$$Im(T) \supseteq Im(T^2)$$

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

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Assuming that T is a linear map from some vector space V to itself... $$\ker{T} \subseteq \ker{T^n}$$ (where n is any natural number) is easy to prove. It's obvious for n = 1. Suppose it's true for n = k. If $x \in \ker{T}$, then $x \in \ker{T^k}$, by the induction hypothesis. But then $T^{k+1}(x) = T(T^{k}(x)) = T(0) = 0$ (by the linearity of T), so that $x \in \ker{T^{k+1}}$. I think the thing about Im(T) could be done without induction.
 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...

Recognitions:
Homework Help

## Image and kernel of T^n

Note that for a linear operator T on a vector space V, we have

$$Im(T^2) = T(T(V)) = T(Im(T))$$

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

$$Im(T^{n+1}) = T^{n+1}(V) = T^n(T(V)) = T^n(Im(T))$$

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

$$Im(T) \subseteq Im(T^2) \subseteq \dots \subseteq Im(T^n) \subseteq \dots$$

which is what you wanted, I suppose.