Inverse of an endomorphism Let $a\in \mathbb{R}\backslash \{0\}$, $n \in \mathbb{N}$ and let $P_n$ be the vector space of polynomial functions of degree less than or equal to $n$. We define on $P_n$ the endomorphism $v$ such that $v(f)=f'-af$. Express $v^{-1}$ using $v, a$ and $I=id_{E_n}$. I managed to show that $v$ is a bijection (so $v^{-1}$ exists). Do you have an idea for the expression of $v^{-1}$?
 A: Consider the operator $g \colon P_n \rightarrow P_n$ given by $g(f) = f'$. Note that $g$ is nilpotent of order $n + 1$. To find $(g - aI)^{-1}$, we can use the geometric series which gives us the inverse of $(1 - x)$ and plug in $g$. More explicitly, we have
$$ (g - aI)^{-1} = \left( (-a) \left( I - \frac{g}{a}  \right) \right)^{-1} = -\frac{1}{a} \left( I - \frac{g}{a} \right)^{-1} = -\frac{1}{a} \sum_{k=0}^{\infty} \left( -\frac{g}{a} \right)^k = \sum_{k=0}^n \frac{(-1)^{k+1}}{a^{k+1}} g^k. $$
This is a formal calculation but now you can just verify that the right hand side indeed gives you the inverse of $g - aI$.
A: Let's look at what happens for $n=2$: if $f_1(x)=1$, $f_2(x)=x$, $f_3(x)=x^2$, then
$$
v(f_1)=-a
\qquad
v(f_2)=1-ax
\qquad
v(f_3)=2x-ax^2
$$
so the matrix of the endomorphism with respect to the basis $\{v_1,v_2,v_3\}$ is
$$
\begin{bmatrix}
-a & 1 & 0\\
0 & -a & 2\\
0 & 0 & -a
\end{bmatrix}
$$
and it is not difficult to show that the general form for the $n$ case is
$$
\begin{bmatrix}
-a & 1 & 0 &  0 & \dots & 0\\
0 & -a & 2 & 0 & \dots & 0\\
0 & 0 & -a & 3 & \dots & 0 \\
\dots \\
0 & 0 & 0 & 0 & \dots & n\\
0 & 0 & 0 & 0 & \dots & -a\\
\end{bmatrix}
$$
The matrix is clearly invertible. and has the form
$$
-aI_{n+1}+B
$$
and the matrix $B$ is nilpotent: $B^{n+1}=0$. Set $c=a^{-1}$, so we want to find the inverse of $cB-I_{n+1}$: since
$$
(I_{n+1}+cB+c^2B^2+\dots+c^nB^n)(cB-I_{n+1})=I_{n+1}
$$
the required inverse is
$$
a(I_{n+1}+cB+c^2B^2+\dots+c^nB^n)
$$
You can also use the fact that
$$
v^{-1}(g)=e^{ax}\int e^{-ax}g(x)\,dx
$$
(we have to choose the unique polynomial antiderivative).
For instance, if $g(x)=1$, we have
$$
v^{-1}(1)=e^{ax}\int e^{-ax}\,dx=-\frac{1}{a}
$$
(the unique polynomial antiderivative). For $g(x)=x$ we need an integration by parts and so on.
There is another way to express the inverse. The transpose is lower triangular, so it can be written as the product of elementary matrices as
$$
E_1(-a)E_{21}(1)E_2(-a)E_{32}{2}\dotsm E_{n}(-a)E_{n+1,n}(n)E_{n+1}(-a)
$$
so the inverse of the transpose is
$$
E_{n+1}(-c)E_{n+1,n}(-n)E_{n}(-c)\dotsm E_{32}(-2)E_2(-c)E_{21}(-1)E_1(-c)
$$
and the required inverse is the transpose of the last one, that is,
$$
E_1(-c)E_{12}(-1)E_2(-c)E_{23}(-2)\dotsm E_n(-c)E_{n,n+1}(-n)E_{n+1}(-c)
$$
where


*

*$E_i(k)$ denotes the matrix coinciding with the identity except at place $(i,i)$ where it has $k$;

*$E_{ij}(h)$ denotes the matrix coinciding with the identity except at place $(i,j)$ where it has $h$.

