Let $V$ be the vector space of real polynomials in $x$ of degree at most $d$ where d>0. Are the following linear mappings of $V$ into itself diagonalisable?
$$(i) \qquad T_1: f(x)\longmapsto x\frac{df}{dx}$$
and
$$(ii)\qquad T_2:f(x)\longmapsto \frac{df}{dx}.$$
I think about the eigenvalue but how to get the eigenvalue from the information?
Observe that $T_1(x^k)=kx^k$, for all $k=0,\ldots,d$. Hence, $\{1,x,\ldots,x^d\}$ is basis of $V$ consisting of eigenvectors of $T_1$. Hence $T_1$ is diagonalisable.
Clearly $T_2^{d+1}f=f^{(d+1)}=0$, for all $f$, but $T_2^d (x^d)= d!\ne 0$. This means that the minimal polynomial of $T_2$ is $m(x)=x^{d+1}$ and hence $T_2$ is not diagonalisable - Here we use the fact that the minimal polynomial of diagonalisable matrix has simple roots.