Can we deduce the characteristic polynomial for this matrix? Given a square $n \times n$ matrix $A$ that satisfies $$\sum\limits_{k=0}^n a_k A^k = 0$$ for some coefficients $a_0, a_1, \dots, a_n,$ can we deduce that its characteristic polynomial is $\sum\limits_{k=0}^n a_k x^k$? 
 A: Take matrix
$$
A = 
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2 \\
\end{array}
\right)
$$
Then
$$  A^4 - 11 A^3 + 41 A^2 - 61 A + 30 I = 0$$
However, the characteristic polynomial is
$$  x^4 - 6x^3 + 13x^2 - 12x + 4 $$
and the minimal polynomial is
$$ x^2 - 3 x + 2 $$
A: The zero matrix satisfies every homogeneous polynomial, but clearly not every homogeneous polynomial is the characteristic polynomial of the zero matrix. 
A: Let $m$ be the minimal polynomial and $p$ be the characteristic polynomial of $A$. Then $m(x)\mid p(x)$ and $m(x)\mid q(x)$, where $q(x)=\sum a_kx^k$. The question is whether or not this actually implies that $p(x)=kq(x)$ for a constant (nonzero) factor $k$, which is clearly untrue. In particular, the polynomial $p(x)/m(x)$ can be switched out for anything else and multiplying back by $m(x)$ will give you another multiple of $m(x)$ with the same degree but is different from a constant multiple of $p(x)$.
The result however is true if we assume that $m(x)$ has degree $n$, since then $p(x)$ and $q(x)$ will both just be a constant multiple of $m(x)$.
A: The answer is no.
You are given a degree $n$ polynomial $p(x)$ that an $n\times n$ matrix $A$ satisfies. This is not enough information to find the characteristic polynomial $c_A(x)$, although you will be able to narrow it down to finitely many possibilities.
Let's look at an example to see why. Suppose your friend picks the matrix $A$. Suppose he doesn't tell you $A$, but he does tell you $p(x)$ and asks you to guess $c_A(x)$. Suppose your friend picked
$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
and told you that $p(x)=x^3-6x^2+11x-6$.
You could then reason that $p(x)=(x-1)(x-2)(x-3)$. Which would mean that the minimal polynomial $m_A(x)$ must be one of the following: $(x-1)$, $(x-2)$, $(x-3)$, $(x-1)(x-2)$, $(x-1)(x-3)$, $(x-2)(x-3)$, $(x-1)(x-2)(x-3)$.
Hence the characteristic polynomial $c_A(x)$ must be one of the following: $(x-1)^3$, $(x-2)^3$, $(x-3)^3$, $(x-1)^2(x-2)$, $(x-1)(x-2)^2$, $(x-1)^2(x-3)$, $(x-1)(x-3)^2$, $(x-2)^2(x-3)$, $(x-2)(x-3)^2$, $(x-1)(x-2)(x-3)$.
Since your friend picked
$$A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
the characteristic polynomial is $c_A(x)=(x-1)^2(x-2)$, but you can't prove that, because for all you know he may have picked
$$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$
which also satisfies $p(x)=x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$.
