Find a vector $a$ such that matrix Aa have geometric multiplicity 3

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let $a=\begin{bmatrix} a0& a1\cdots &an \end{bmatrix} \in \mathbb R^{n+1}$ be some fixed vector, and $A_a:P^n\to P^n$ be the linear transformation:

$A_a(p)=a_0p+a_1\frac{d}{dx}p+\cdots +a_n\frac{d^n}{dx^n}p$,

where $P^n$ is space of real polynomial which degree is not bigger then n.

a)Find the matrix of the linear transformation $A_a$ in the standard basis.

$$ A_a=\begin{bmatrix} a_0 & a_1 & 2a_2& 6a_3& \cdots &n!a_n \\ 0& a_0 & 2a_1 & 6a_2 & \cdots &n!a_{n-1} \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ 0& 0& 0& 0&0& a_0 \end{bmatrix} $$

b)Find a vector $a\in \mathbb R^{n+1}$ for which eigenvalue(only one eigenvalue) of linear transformation has geometric multiplicity 3, then on the same way find example, and show that for every $k\in{1,2,\cdots,n+1}$ exist vector $a^k \in \mathbb R^{n+1}$, such that only that one eigenvalue of linear transformation have geometric multiplicity $k$.

For b i have no idea, because I know in start that I have geometric multiplicity 1 because one eigenvalue, but i have no idea how to get 3 or more because it look that everything is zero, or I write matrix in wrong way, can someone help me?