# Why does the largest Jordan block determine the degree for that factor in the minimal polynomial?

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?

• 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... – J. M. is a poor mathematician Nov 16 '11 at 7:41

• 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 $p(SAS^{-1}) = 0$.
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$~J$.