= Solution
The <polynomials> $t^c$ and $(t-1)^d$ are <coprime>. <Bézout's identity> gives <polynomials> $u,v$ with $u(t)t^c+v(t)(t-1)^d=1$. Set
$$
V_0=\ker T^c,\qquad V_1=\ker(T-I)^d.
$$
For every $x$, the decomposition
$$
x=v(T)(T-I)^dx+u(T)T^cx
$$
has first summand in $V_0$ and second in $V_1$, since $T^c(T-I)^d=0$ and <polynomial> expressions in $T$ commute. If $x$ belongs to both <kernel of a linear map>[kernels], the same Bézout identity gives $x=0$. Thus $V=V_0\oplus V_1$. Both <vector subspaces> are <invariant> under $T$.
On $V_0$, the restriction of $T$ is <nilpotent operator>[nilpotent]; on $V_1$, the restriction of $T-I$ is <nilpotent operator>[nilpotent]. Apply (b) separately, then concatenate the two <bases>. This yields
$$
\boxed{[T]=\begin{pmatrix}A&0\\0&I+N\end{pmatrix},}
$$
where $A,N$ are <strictly upper triangular matrix>[strictly upper triangular]. Empty blocks are allowed if one subspace is zero. In particular $C=I+N$ has precisely the required diagonal.
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