Linear Transformation problem. .

kNYsJakE
Messages
25
Reaction score
0

Homework Statement


Let \mathrm{V} be a vector space. Determine all linear transformations \mathrm{T}:V\rightarrow V such that \mathrm{T}=\mathrm{T}^2.


Homework Equations


Hint was given and it was like this:
Note that x=\mathrm{T}(x)+(x-\mathrm{T}(x)) for every x in V, and show that V=\{y:\mathrm{T}(y)=y\}\oplus\mathrm{N}(T)



The Attempt at a Solution


I tried to calculate T(x) and T^2(x) using x=T(x)+(x-T(x)) and put T and T^2 equal. But, I do not think this is how to solve this problem... =(
 
Physics news on Phys.org
Each x can be written as x = T(x) + (x-T(x)) so note that:

T(T(x)) = T(x) so T(x) is in the set {y:T(y)=y}

And T(x-T(x)) = T(x) - T2(x) = T(x) - T(x) = 0 so x-T(x) is in N(T).

So given a T, we can represent V as the direct sum of the kernel of T and the image of T. The opposite question is, given V as the direct sum of two subspaces, can we find a T such that one is the kernel and the other is the image?
 
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