# Linear Transformation On V.

1. Jan 11, 2010

### nlews

Hello,

I am working through some examples for revision purposes and am pondering over this question so would appreciate any help I could receive.

I would like to prove that if T is a linear transformation on V such that T^2 = T, and I is the identity transformation on V,

i)Ker(T) = im(I-T) and ker(I-T) =im (T)
ii) kerT n imT = {0}
ii)and that every vEV can be uniquily expressed in the form v=u+w where u E kerT and w E imT

Attempts:
i)
I am unsure how to begin this question.

ii) We know that both ker(T) and ker (T^2) will have the same dimension, therefore it follows that we have equality, kerT = ker(T^2)
Suppose v E Ker(T) n Im(T) then Tv= 0 and so (T^2)v =0, but then w E ker(T^2). Therefore, we have that w E ker T, but then v = T(w) = 0, therefore ker(T)n im(T) = {0}
I think this is quite confused, I cannot see the correct logic but will keep trying to come up with a clearer proof, any help would be good aswell.

iii) Suppose there are u,u' E kerT and w,w' E imT such that v= u+w and v = u'+w'
then these equations imply that u-u'=w-w' E kerT n imT = {0} (from part ii)

Thank you for your help in advance

2. Jan 11, 2010

### rochfor1

i) You know that 0 = T - T^2 = T(I-T), so im(I-T) is a subset of ker T. The reverse inclusion is even easier and I'll leave it to you. The proof that im T = ker(I-T) is quite similar.

ii) Your proof is pretty good, but you neglected to say that w is in V such that Tw = v, so that might be causing a little confusion.

iii) You've given a good proof that every such decomposition is unique, but you haven't said why such a decomposition must exist. To do so, consider the identity I = (I-T)+T.