Cayley-Hamilton equation for a given matrix $A$ and other matrices We know  that "every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic polynomial" i.e.  
$p(A)=0$
My question is:


*

*what other matrices are satisfying this particular equation $p(A)=0$


Evidently it is satisfied  by any matrix which is generated from $A$ through a change of basis and also by some more simply generated matrices, for example $A^T$ or $\lambda_i{I}$, where $\lambda_i $ is eigenvalue for matrix $A$.
But maybe it is possible to find a general form of the matrix which is satisfying characteristic equation generated for a matrix A?
 A: Suppose we work over fields. $p(T) = 0$ if and only if the minimal polynomial $m_T$ of $T$ divides $p$, i.e. there is a polynomial $q\in\mathbb{F}[\lambda]$ with $p = q\cdot m_T$. So the question comes down to: which linear operators have a minimal polynomial that divides $p$? Which begs the question, given a polynomial $m$, what linear operators have $m$ as their minimal polynomial?
For each factor $q$ of $p$, one can find at least one matrix whose minimal polynomial is $q$, namely the companion matrix to $q$.
If $\mathbb{F}$ is a field where polynomials split into linear factors (e.g. $\mathbb{C}$), then the Jordan form is well-defined for all linear operators. The minimal polynomial in this case also factors. The minimal polynomial restricts the Jordan form of the operator in the following way: the degree of each linear factor corresponds to the size of the largest Jordan block. So for example, if $(\lambda-1)^3$ is a factor of $m_T(\lambda)$, then the Jordan canonical form of $T$ is a matrix with at least one $3\times 3$ Jordan block of eigenvalue $1$. These are the only restrictions placed on the Jordan form by the minimal polynomial.
Since there are only finitely many possible factors of $p$, if $p(T) = 0$ then there are only finitely many possible minimal polynomials for $T$. If we also fix the $k$ so that $T$ is a linear mapping $\mathbb{F}^k\to\mathbb{F}^k$, then for each choice of minimal polynomial there are only finitely many possible Jordan forms for $T$. You can thus use the Jordan forms to explicitly classify all operators that annihilate $p$.
The above discussion relies heavily on the fact that $\mathbb{F}[\lambda]$ is a principal ideal domain (namely so that $m_T$ is defined). If we take coefficients in a commutative ring that is not a field then the above will likely fail.
A: All those matrices which have the same characteristic polynomial as that of $A$ will surely satisfy the equation $p(A)=0$ viz. $A^T$,any matrix similar to $A$ as you have pointed out.
A matrix $B$ will satisfy the characteristic polynomial of $A\iff$  the minimal polynomial of $B$  divides the characteristic polynomial of $A$.
Let the characteristic polynomial of $A$ be $\lambda_A(x)$ and the minimal polynomial of $B$ be $m_B(x)$.
The minimal polynomial of $B$  divides the characteristic polynomial of $A\implies \lambda_A(x)=m_B(x)q(x)\implies m_B(A)q(A)=0$
Since we are in a field $m_B(A)q(A)=0\implies m_B(A)=0$.
Conversely suppose that the minimal polynomial of $B$ satisfies the characteristic polynomial of $A\implies m_B(A)=0$.
To show : $m_B(x)\text{divides } \lambda_A(x)$.
Proof:The characteristic polynomial of $A,\lambda_A(x)$ annihilates $B$  and the minimal polynomial of $B,m_B(x)$ divides any annihilating polynomial of $B $ and hence $m_B(x)\text{divides } \lambda_A(x)\text{Q.E.D.}$
