Determinant of map $p(x) \mapsto (Tp)(x)=a_n+a_{n-1}x+ \ldots +a_0x^n$ 
Let $V$ be the vector space of polynomial $\mathbb{R}$ of degree less than or equal to $n$. For $p(x)=a_0+a_1x+ \ldots +a_nx^n$ in $V$. Define a Linear Transformation $T:V \to V$ by $(Tp)(x)=a_n+a_{n-1}x+ \ldots +a_0x^n$ Then

*

*$T$ is one to one

*$T$ is onto

*$T$ is invertible

*$\det T = 1$ or $-1$.


If I take standard basis then I am getting all options correct but if I take $p(x)=3x+3$ then getting last option wrong as value of determinant coming $9$. I know rest of options are right.
 A: Sketch:
The matrix of this linear transformation, in the standard basis for $\mathbf R_n[x]$ is simply the antidiagonal matrix:
$$\begin{bmatrix}0&0&0&\dots&0&1\\
0&0&0&\dots&1&0\\&\vdots&&&\vdots\\
0&1&0&\dots&0&0\\
1&0&0&\dots&0&0\end{bmatrix}$$
Now use the definition of the determinant:
$$\sum_{\sigma\in S_{n+1}}\varepsilon(\sigma)a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n+1\sigma(n+1)}$$
The only non-zero product in this formula is $a_{1\, n+1}a_{2\,n}\dots a_{n+1\,1}$, which corresponds to the permutation $$ \begin{pmatrix}1&2&\dots&n+1\\n+1&n&\dots&1\end{pmatrix}=(1,n+1)(2,n)\dots (p,p+1)$$
as a product of transpositions, whether $n=2p$ ot $n=2p-1$. Hence the determinant is the signature of this permutation; i.e. $\;\color{red}{(-1)^p}$.
A: If you apply twice the map, you get back to where you started: $T(T(p))=p$. By Binet's formula,
$$
(\det T)^2=1
$$
because the identity has determinant $1$. Therefore $\det T=1$ or $\det T=-1$.
Since $\det T\ne0$, the map is invertible, so onto and one-to-one.
