Let $n \in \mathbb{N}$. We define:
$f_n:\mathbb{R}[x]_{\leq n} \rightarrow \mathbb{R}[x]_{\leq n}, p\mapsto(p \cdot x)' $, where $q'$ is the derivation of a math. polynomial $q \in \mathbb{R}[x]$.
Show that $f_n$ is an endomorphism. Calculate the determinant of $f_5$.
My solution so far:
endomorphism:
$f$ is a transformation from $\mathbb{R}[x]_{\leq n}$ to $\mathbb{R}[x]_{\leq n}$.
need to show: $f(p_1(x)+\lambda p_2(x))=f(p_1(x) + \lambda f(p_2(x)).$
Let $p_1(x) = a_nx^n+...+a_1x+a_0,~ p_2(x) = b_nx^n+...+b_1x+b_0$ and $\lambda \in \mathbb{R}$
$f(p_1(x)+\lambda p_2(x))=f(a_nx^n+...+a_1x+a_0+\lambda(b_nx^n+...+b_1x+b_0))=f(a_n+\lambda b_n)x^n+...(a_1+\lambda b_1)x+(a_0+\lambda b_0))= (n+1)(a_n+\lambda b_n)x^{n}+...+2(a_1+b_1)x+(a_0+b_0)=(n+1)(a_n)x^{n}+\lambda((n+1)b_n)x^{n}+...+(a_0+\lambda b_0)=(n+1)(a_n)x^{n}+...+a_0+ \lambda((n+1)b_nx^{n}+...+b_0)=f(p_1(x) + \lambda f(p_2(x))$
$det(f_5)=det \begin{pmatrix}6 & 0 & 0 & 0 & 0 & 0\\0 & 5 & 0 & 0 & 0 & 0\\0 & 0 & 4 & 0 & 0 & 0\\0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix} =6!$
Is this corrrect?
I don't think this is correct: the endomorphism here is not the differential one but the differential one times $\;x\;$ , and thus $$p\in\ker f_n\iff (xp)'=0\iff xp=k=\text{constant}\iff p=0$$ so $\;f_n\;$ is an isomorphism and thus its determinant is non zero. In fact, $\;\det f_n=(n+1)!\;$ ...