Find all $3 \times 3$ matrices $A$ satisfying $A^3 + 3A = I$. 
Find all $3 \times 3$ matrices $A$ such that $A^3+3A=I$, where $I$ is the identity. 

I already know one matrix satisfying this equation, which is ((0,0,1),(1,0,-3),(0,1,0)), and I want to show that all matrices satisfying the equation are conjugate to this matrix. I know that all matrices satisfying this equation have characteristic polynomial x^3+3x-1 and I guess that this is also the minimal polynomial for all real matrices satisfying it but I am unable to show it. I think that all matrices having same characteristic and minimal polynomials are the same, which I can use to prove the above, but am unable to prove this result too.
 A: The polynomial $x^3 + 3x - 1$ has three distinct roots: one real, and two conjugate complex roots. If you assume that it is the minimal polynomial for a (let's say) complex matrix $A$, then the three roots of the polynomial must be the three eigenvalues of $A$, each of which are distinct, meaning that $A$ is diagonalisable. That is, $A$ is similar to the diagonal matrix containing the roots of $x^3 + 3x - 1 = 0$ in the diagonal. Thus, any two such matrices $A$ and $B$ must be similar to each other, exactly as you want.
Note that I assumed that the polynomial $x^3 + 3x - 1$ is minimal. If we relax this condition, then we will obtain other such matrices. For example, if $\lambda$ is a root of $x^3 + 3x - 1$, then $A = \lambda I$ will be a matrix, not similar to the previous matrices, that will satisfy the given polynomial equation.
Since $x^3 + 3x - 1$ is square-free (i.e. it has distinct roots), we know that the minimal polynomial of $A$ must be a (square-free) factor of this polynomial, and hence $A$ is diagonalisable with eigenvalues taken from the roots of $x^3 + 3x - 1$. You'll have to consider the three roots $\lambda_1, \lambda_2, \lambda_3$, and all possibility of repeated eigenvalues taken from this list of three roots. These will form the similarity classes of matrices which satisfy this polynomial equation.
