diagonalisation and basis? What are the conditions for something to be diagonalisable with regard to basis?
I am trying to do this question:

Let $V$ be a real $n$-dimensional vector space, and $T:V \rightarrow V$ be a linear mapping.  Show that if $\lambda$ is the only eigenvalue of $T$ and $T$ is diagonalisable then $T=\lambda I$.
Now let $V$ be the vector space of real polynomials in $x$ of degree at most $d$ where $d>0$.  Which of the following linear mappings of $V$ into itself are diagonalisable?
  
  
*
  
*$T_1:f(x) \mapsto x\tfrac{df}{dx}$
  
*$T_2:f(x) \mapsto \tfrac{df}{dx}$
  
*$T_3:f(x) \mapsto f(x+1)$
  
*$T_4:f(x) \mapsto f(-x)$.
  

I found matrix of $T_1, T_2,\dotsc$ with respect to basis $B = \{1, x^2, x^3, ... , x^d\}$.
But I dont know how to tell which ones are diagonalisable?
Thanks
 A: In general the most definitive way to decide whether a linear operator is diagonalisable is to compute its minimal polynomial and see if it splits into distinct factors $X-a_i$ (that is a necessary and sufficient condition for being diagonalisable). But in this exercise it is much easier to just ask what an eigenvector would look like. For instance for the operations that substitutes $X+1$ for $X$ in a polynomial: $P[X]\mapsto P[X+1]$, it is hard to imagine that this multiplies $P$ by a scalar other than$~1$ (and you can prove that it is impossible by looking at the leading nonzero coefficient of$~P$). Moreover eigenvectors for $\lambda=1$ are polynomials unchanged under the substitution; again this seems to be hard to achieve, except for the case of constant polynomials. But given that $\lambda=1$ is the only possible eigenvalue, it is already clear that being diagonalisable could only be the case if the operation is the identity, and this is clearly not the case if (unless the space is limited to polynomials of degree${}\leq0$).
