Let's say I have the characteristic polynomial of an operator:


Wouldn't then the minimal polynomial be exactly:

$$q(z)=(z-\lambda_1)(z-\lambda_2)\dots(z-\lambda_n)\text{ ?}$$

It seems to me like it should be, since it is the polynomial of least power that has all the eigenvalues as roots. Still, people seem to be using more complicated processes to calculate the minimal polynomial. If this holds, won't it be trivial given a characteristic polynomial? The characteristic polynomial can itself be obtained from an upper triangular matrix of the operator.

Or is it simply the case that there exists a simpler polynomial than the one having the eigenvalues as roots, which still gives $q(T)=0$?

  • $\begingroup$ No. It is the smallest degree monic polynomial $\psi$ such that $\psi(A) = 0$. Let $A$ be a Jordan block and compute the minimal polynomial. $\endgroup$ – copper.hat Jul 31 '14 at 17:51
  • $\begingroup$ The minimal polynomial will be $(z-\lambda_1)^{k_1}(z-\lambda_2)^{k_2}\dotsc (z-\lambda_n)^{k_n}$ for some exponents $1 \leqslant k_\nu \leqslant j_\nu$. The exponents depend on the normal form, consider $$\begin{pmatrix} \lambda & 1 & 0\\ 0 & \lambda & 1\\ 0 & 0 & \lambda\end{pmatrix}.$$ $\endgroup$ – Daniel Fischer Jul 31 '14 at 17:52
  • $\begingroup$ @user163916 : The polynomial you mention is the minimal polynomial of that set of eigenvalues, but "minimal polynomial" probably was intended, in the context in which you saw it, to mean the minimal polynomial of a matrix. $\endgroup$ – Michael Hardy Jul 31 '14 at 18:18

The minimal polynomial of $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is $z^2$.

copper.hat and Daniel Fischer also posted answers in the comments with more generality.


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