Let $T:\mathbb{R}_n\rightarrow\mathbb{R}_n$ the linear operator defined by $T(p(x))=p(x)+p'(x)$ Calculate $det(T)$
My work:
Let $B=\{1,x,x^2,...,x^n\}$ a basis for $\mathbb{R}_n$.
Then,
$T(1)=1$
$T(x)=x+1$
$T(x^2)=x^2+2x$
.
.
.
$T(x^n)=x^n+nx^{n-1}$
This implies,
$[T_{BB}]=\begin{pmatrix} 1 && 1 && 0 && 0 && ... && 0\\ 0 && 1 && 2 && 0 && ... && 0\\ 0 && 0 && 1 && 3 && ... && 0\\ .\\ .\\ .\\ 0 && 0 && 0 && 0 && ... && n\\ 0 && 0 && 0 && 0 && ... &&1\\ \end{pmatrix}$
As $det(T)=det([T_B])$ and $det([T_B])=det\begin{pmatrix} 1 && 1 && 0 && 0 && ... && 0\\ 0 && 1 && 2 && 0 && ... && 0\\ 0 && 0 && 1 && 3 && ... && 0\\ .\\ .\\ .\\ 0 && 0 && 0 && 0 && ... && n\\ 0 && 0 && 0 && 0 && ... &&1\\ \end{pmatrix}=1$
then $det(T)=1$.
I don't sure of my result. I think i have a mistake. Can someone help me?