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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.

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1 Answer 1

up vote 12 down vote accepted

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.

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This is a very clear example, thank you –  Ch Zh Feb 17 '12 at 12:46
    
You are welcome. –  Calle Feb 17 '12 at 12:50

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