# 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

## Answers and Replies

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.

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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...

AKG
Science Advisor
Homework Helper
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.

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