I have computed (a) to be $-\lambda^3$. I also know that the Cayley-Hamilton theorem states that substituting the matrix A (where A is matrix with p(λ)=det(λI-A) for λ in this polynomial results in the zero matrix p(A)=0. But don't know how this verifies for f. I'm unsure how I find (c).
1 Answer
It is easy to see if you choose the right basis. The obvious choice is $1,x,x^2$. With respect to such basis, the matrix of $f$ is $$ \begin{bmatrix} 0&1&0\\0&0&2\\0&0&0\end{bmatrix}. $$ Then you can easily see that $f^3=0$ (which is the fact that the third derivative of a degree 3 polynomial is zero). And the characteristic polynomial is also the minimal polynomial, because the minimal polynomial divides the characteristic polynomial; but $f^2\ne0$.