Recent content by Tazz01

Edit, I've now completed this question, thanks for your help guys.

I can confirm that I've solved this problem.

The Question: 2 people A, B play a series of independent games. We have the following probabilities: P(A wins a game) = p P(B wins a game) = q = 1 - p Both players begin with X units of money, and in each game the winner takes 1 unit from the other player. The series terminates when...

I've proved i-ii and ii-iii but not yet iii-i. This is something I'm stuck on, can someone advise?

I'm not sure about that Bacle, I will assume that the question has a solution and that ker(T) does equal ker(T^2). Would anyone be able to advise how I could go about implying (ii) from (i)? Do I have to show that (i) holds for ker(T^2)? And then use rank-nullity from ii - iii?

I think I figured out how to show ker(T)=ker(T^{2}). Say for x \in ker(T), then we have for T(T(x)): T(T(x)) = T(0) <--- Now this is the zero vector inside the brackets Can we pull this out as a contant, and use any vector v \in R^{3}: T(0) = T(0v) = 0T(v) = 0 Therefore, we have...

If x is in ker(T), this means that T(x)=0. If we put this same x into T(T(x)), we get T(0) - but we don't know what T(0) equals... I'm still lost on this, how can we approach the first part, showing that i ==> ii?

Are the domain and the codomain for T^{2} the same as they were for T? e.g. T^{2}:R^{3}->R^{3} This would mean that: ker(T^{2})={x\inR^{3} : T^{2}(x)=0} Similarly for the image. It seems that to prove (ii) and (iii) are the same, we can use the rank-nullity theorem, but no idea for...

Homework Statement T : R^{3} -> R^{3} is a linear transformation. We need to prove the equivalence of the three below statements. i) R^{3} = ker(T) \oplus im(T); ii) ker(T) = ker(T^{2}); iii) im(T) = im(T^{2}). Homework Equations R^{3} = ker(T) \oplus im(T), if for all v \in...