= Solution
Relative to the coordinate basis, the <linear map> has matrix
$$
F=\begin{pmatrix}1&3&-1\\0&2&1\\0&-4&-1\end{pmatrix}.
$$
Expanding $\det(tI-F)$ in its first column gives the monic <characteristic polynomial>
$$
\boxed{\chi_f(t)=(t-1)(t^2-t+2)}.
$$
Its <eigenvalues> are $1$ and $(1\pm i\sqrt7)/2$, three distinct complex numbers. <Eigenvectors> corresponding to distinct <eigenvalues> are linearly independent, so they form a basis of $\mathbb C^3$ and \b[$f$ is diagonalizable].
Every annihilating <polynomial> must vanish at all three <eigenvalues>: applying $p(f)$ to a corresponding <eigenvector> gives $p(\lambda)v$. Conversely, in the eigenbasis the product of the three distinct linear factors annihilates $f$. Hence the <minimal polynomial> is
$$
\boxed{m_f(t)=(t-1)(t^2-t+2)}.
$$
If $af+bf^2=0$ with scalars $a,b$ not both zero, the nonzero <polynomial> $at+bt^2$ of degree at most two would annihilate $f$, contradicting this minimal degree of three. Thus \b[$f$ and $f^2$ are linearly independent <endomorphisms>].
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