Let $A$ be a square matrix, so $A$ has some Jordan Normal form. Then $A$ has a minimal polynomial, say $m(X)=\prod_{i=1}^k (t-\lambda_i)^{m_i}$.

Wikipedia says

The factors of the minimal polynomial $m$ are the elementary divisors of the largest degree corresponding to distinct eigenvalues.

So $m_i$ is the size of the largest Jordan block of $\lambda_i$. Why is this exactly?

  • 1
    $\begingroup$ You know the minimal polynomial of a single Jordan block, no? See what happens if you try to find the minimal polynomial of a block-diagonal matrix with just one eigenvalue, but with Jordan blocks of various sizes... $\endgroup$ Nov 16, 2011 at 7:41

2 Answers 2



  • a single Jordan block $B$ of size $m$ with eigenvalue $\lambda$ has $(B - \lambda I)^m = 0$ but $(B - \lambda I)^{m-1} \ne 0$,

  • if a square matrix $A$ has blocks $B_1, \ldots, B_k$ along the diagonal and $0$'s everywhere else, and $p$ is any polynomial, $p(A)$ has blocks $p(B_1), \ldots, p(B_k)$ along the diagonal and $0$'s everywhere else

  • and if $A$ and $S$ are square matrices of the same size with $S$ invertible, and $p$ is any polynomial, $p(S A S^{-1}) = S\ p(A) S^{-1}$; in particular $p(A) = 0$ if and only if $p(SAS^{-1}) = 0$.

  • $\begingroup$ Could you elaborate a little on how you are using the last fact to conclude the result? $\endgroup$ Nov 20, 2020 at 4:07
  • $\begingroup$ By the way, is there a name for this $m$ (= the size of the largest Jordan block of $\lambda$ = the multiplicity of $\lambda$ in the minimal polynomial of $A$) ? $\endgroup$ Nov 29, 2021 at 9:24

By construction the Jordan block $J$ for$~\lambda_i$of size $m_i$ contains a vector $v$ such that the vectors $v$, $(A-\lambda_iI)(v)$, ... $(A-\lambda_iI)^{m_i-1}(v)$ form a basis of$~J$, and with $(A-\lambda_iI)^{m_i}(v)=\vec0$. So certainly the $m_i$-th power of $A-\lambda_iI$ is the smallest one that will annihilate this Jordan block$~J$. At le same time it will annihilate all other (smaller) Jordan blocks for$~\lambda_i$. Any other factors in the product forming $m(A)$ act in an invertible way on the generalized eigenspace $V_{\lambda_i}$ for $\lambda_i$ (the kernel of the restriction of such a factor to$~V_{\lambda_i}$ is zero), and in particular on$~J$, so their presence makes no difference for annihilating$~J$.

Therefore, if you take for every eigenvalue as exponent the maximum size of a corresponding Jordan block, you do annihilate all generalized eigenspaces. Since you assumed that these generalised eigenspaces span everything (i.e., there exists a Jordan normal form, which means the minimal (and characteristic) polynomial is split), you have your minimal polynomial.


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.