Is T^n Linear When T is Linear?

cristina89
Messages
29
Reaction score
0

Homework Statement


If T is a linear transformation, proof that Tn is a linear transformation (with nEN).


Homework Equations


I know that T is a linear application if:
T(u+v) = T(u) + T(v)
T(au) = aT(u)

The Attempt at a Solution


Actually I don't know how to start using these two affirmations. Can anyone help me with it?
I know how to do this when it has numbers, but then it comes to these kind of proofs, I don't know how to do this.
 
Physics news on Phys.org
Start with T2. Is it true that T2(u+v)=T2(u)+T2(v)? Note that T2(u+v) means T(T(u+v)).

ehild
 
You can do the general proof "by induction".
 
I'm trying to solve it by induction.

For n = 1 ok.

Assuming that's ok for n = k.

For n = k+1

I don't know if I'm doing it right in this part:

Tk+1 = Tk.T(u+v) = Tk.(T(u+v)) = Tk(T(u)) + Tk(T(v)). Can I just afirm that's ok since T(u+v) is an application and Tk is an application too?
 
I would have put in one more step. Tk(T(u+ v))= Tk(T(u)+ T(v)), using the "given" fact that T is linear, and then "= Tk(T(u))+ Tk(T(v))" using the "induction hypothesis" that Tk is linear.

And, of course, you now need to prove that Tn(au)= aTn(u) but that can be done the same way.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top