Re: transformation question

$\displaystyle T^2-T=-I$. Now see if the kernel has any vectors different from zero.

Re: transformation question

Slight variation: $\displaystyle T- T^2= T(I- T)= (I- T)T= I$

Re: transformation question

Can you commute T and I-T?

Re: transformation question

Yes, of course: $\displaystyle T- T^2= I*T- T*T= T*I- T*T$. Any linear operator from a vector space to itself commutes with itself and the identity.

Re: transformation question

Quote:

Originally Posted by

**HallsofIvy** Slight variation: $\displaystyle T- T^2= T(I- T)= (I- T)T= I$

Quote:

Originally Posted by

**ModusPonens** Can you commute T and I-T?

Moreover, judging from the level of the question one would guess that one can assume that the vector space is finite dimensional from where $\displaystyle T(\mathbf{1}-T)=\mathbf{1}$ implies $\displaystyle T$ is surjective from where (by finite dimensionality) bijectivity follows.