= Solution
{c}
In the <monomial basis> $\mathcal B=(1,x,x^2,x^3)$ of the <complex vector space> $V=\mathbb C[x]_{\leq3}$, the <binomial theorem> gives
$$
[T]_{\mathcal B}=
\begin{pmatrix}
1&1&1&1\\
0&1&2&3\\
0&0&1&3\\
0&0&0&1
\end{pmatrix}.
$$
The <characteristic polynomial> of this <upper triangular matrix> is the product of $t$ minus its <diagonal entries>, so $\chi_T(t)=(t-1)^4$.
For the <minimal polynomial>, set $\Delta=T-I$, the <forward difference> with step one. For a nonconstant <polynomial> of <polynomial degree> $d$, the <binomial theorem> shows that $f(x+1)-f(x)$ has <polynomial degree> $d-1$. Constant <polynomials> have zero <forward difference>. Consequently $\Delta^4=0$ on $V$. To prove that the <nilpotency index> is exactly four, compute
$$
\Delta x^3=3x^2+3x+1,\qquad
\Delta^2x^3=6x+6,\qquad
\Delta^3x^3=6\ne0.
$$
It follows that
$$
\boxed{\chi_T(t)=(t-1)^4,\qquad m_T(t)=(t-1)^4.}
$$
If $T$ were <diagonalizable>, its only <eigenvalue> would be $1$, and its <diagonal matrix> in an <eigenbasis> would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus \b[$T$ is not <diagonalizable>]. Equivalently, its <minimal polynomial> has a <repeated root>.
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