Let $V$ be a $K$-vector space with basis $B = (v_1,\ v_2,\ v_3,\ v_4)$ and $f \in End(V)$ with $$f(v_i) = v_{i+1},\ \ \ f(v_4) := v_1$$ Calculate the characteristic polynomial $P_f(t)$ of $f$.
I know that $$P_f(t) = P_A(t)$$
where $A := M^B_B(f)$ the transformation matrix. Then since the matrix is triangular M_B^B(f) = \begin{pmatrix} B & * \\ 0 & B' \end{pmatrix} we would have $$P_f(t) = P_B(t)P_{B'}(t)$$ Performing Gaussian Elimination over the columns of $B'$ - 4 swaps, would transform $B'$ into $B$ and not affect the sign of the determinant, so $$P_f(t) = P_B(t)^2$$ $$P_f(t) = (t^4 - a_1t^3 + a_2t^2 + a_1t + a_4)^2$$
And how to proceed in order to fully calculate $P_f(t)$? Was so far correct? Any remarks?
Thanks.
HINT: What is $f^4(v_i)$? What does that tell you about the characteristic polynomial of $f$?
Alternatively you could simply write out $f$ on the basis $B$; it is simply $$M_B^B(f)=\begin{pmatrix}0&0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix},$$ so computing the characteristic polynomial is routine.