Do the algebraic and geometric multiplicities determine the minimal polynomial? Let $T$ denote some linear transformation of a finite-dimensional space $V$ (say, over $\mathbb{C}$).
Suppose we know the eigenvalues $\{\lambda_i\}_i$ and their associated algebraic multiplicities $\{d_i\}_i$ and geometric multiplicities $\{r_i\}_i$ of $T$, can we determine the minimal polynomial of $T$ via these informations?
If the answer is no, is there a nice way to produce different linear transformations with same eigenvalues and associated algebraic and geometric multiplicities?

Some backgraoud: It is well-known that for a given linear transformation, the minimal polynomial divides the characteristic polynomial: $m_T|p_T$. And I find in a paper proved that
$$m_T|\prod_i(x-\lambda_i)^{d_i-r_i+1}\ ,\ \ \ \  p_T|m_T\prod_i(x-\lambda_i)^{r_i}$$ 
And then I want to know if there are any better results.

No, the algebraic and geometric multiplicities do not determine the minimal polynomial. Here is a counterexample: Consider the Jordan matrices $J_1, J_2$:
$$J_1 = \left(
\begin{array}{cccc}
 1 & 1 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 1 \\
 0 & 0 & 0 & 1
\end{array}
\right)
~~
J_2 =
\left(
\begin{array}{cccc}
 1 & 1 & 0 & 0 \\
 0 & 1 & 1 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1
\end{array}
\right)
$$
both have only one eigenvalue, namely 1, so they both have algebraic multiplicity 4. They also both have geometric multiplicity 2, since there are 2 Jordan blocks in both matrices (check the Wikipedia article on Jordan normal form for more information). However, they have different minimal polynomials:
$$\begin{align}
m_{J_1}(x) = (x - I)^2 \\
m_{J_2}(x) = (x - I)^3
\end{align}$$
so the algebraic and geometric multiplicities do not determine the minimal polynomial.
