How to calculate the trace and the determinant of a matrix given an annihilating polynomial? 
Find the trace and determinant of the $5 \times 5$ matrix $A$ that satisfies the condition $$ \sum_{k=0}^{5} (k+1)A^k = 0 $$


My attempt relies on the fact that the given condition can be thought as its Cayley-Hamilton polynomial, since it can be written exactly as $P_5(A) = 0$. Considering this fact, I can also say that the characteristic polynomial of the matrix is $P_5(\lambda) = 0$. As check for my attempt, I tried to compute the roots of that polynomial with WolframAlpha, and discovered that the "sum of roots = $-\frac{5}{6}$" and "product of roots = $-\frac{1}{6}$". Those results are indeed the trace and determinant of $A$, as the book give me also the solution, but since I'm not able to find easily these roots (I was't able to find them at all!), I think that my attempt is wrong. Any suggestion?
 A: You need a bit more work to implement what you want. The fact that a polynomial of degree $5$ is zero on a $5\times5$ matrix is not enough to tell that it is the characteristic polynomial. For a simple example consider the $5\times5$ matrix $E_{11}$ (this is the matrix with $1$ at the $1,1$ entry and zero elsewhere). The characteristic polynomial of this matrix is $x^4(x-1)$. But since the minimal polynomial is $x(x-1)$, the polynomials $x^3(x-1)^2$,  $x^2(x-1)^3$, and $x(x-1)^4$ are also degree-five polynomials that annihilate $E_{11}$.
What works in this example to guarantee that your $P_5$ is indeed the characteristic polynomial of $A$ is that $P_5$ has five distinct roots. As mentioned by Macavity in the comments, one can do this by showing that the greatest common divisor between $P_5$ and $P_5'$ is $1$; that would show that $P_5$ has no common roots with its derivative, and thus guarantee that all roots are simple. When the five roots are distinct, you know that $P_5$ is the minimal polynomial, and being of degree $5$ it has to be the characteristic polynomial.
As for how to answer the question, over $\mathbb C$ the (monic version of) the characteristic polynomial is
$$
(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)(x-\lambda_4)(x-\lambda_5).
$$
The independent term of this polynomial is $$(-1)^5\lambda_1\lambda_2\lambda_3\lambda_4\lambda_5=-\det A.$$ And the coefficient of $x^4$ is
$$
-(\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5)=-\operatorname{Tr}(A).
$$
So, since the monic characteristici polynomial of $A$ is
$$
x^5 + \frac56x^4 + \frac46x^3 + \frac12x^2 + \frac13x + \frac16,
$$
it follows that
$$
\operatorname{Tr}(A)=-\frac56,\qquad\qquad\det A=-\frac16.
$$
