11
$\begingroup$

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.

$\endgroup$
25
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.