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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?

Rodrigo de Azevedo
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Boma
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    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. ain't a mathematician Nov 16 '11 at 07:41

2 Answers2

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Because

  • 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 $p(SAS^{-1}) = 0$.

Robert Israel
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    Could you elaborate a little on how you are using the last fact to conclude the result? – stoic-santiago Nov 20 '20 at 04:07
  • 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 $A$) ? – Pietro Majer Nov 29 '21 at 09:24
  • How does the proof rely on the block being the largest one among those corresponding to $\lambda_i$? – mathslover Nov 12 '22 at 16:21
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    @mathslover If $m$ is the size of a Jordan block for $\lambda_i$ but there is another block $B$ of larger size, then $(B-\lambda I)^m \ne 0$. – Robert Israel Nov 13 '22 at 19:46
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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 the 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 $\lambda_i$ ($V_{\lambda_i}$ is stable for the action $(A-\lambda_jI)^k$ of such a factor, so that the latter can be restricted to $V_{\lambda_i}$, and the kernel of the restriction is $V_{\lambda_i}\cap\ker((A-\lambda_jI)^k)=\{0\}$, so the restriction is invertible), 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.

Marc van Leeuwen
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  • "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)", after reading this sentence 3 times, I get it translate to this math operation: $(A-\lambda_j I)^n, v_i \in ker(A-\lambda_i)^{m_i}$ – Kuo Jan 27 '24 at 17:38
  • But I still can not get the meaning of "the kernel of the restriction of such a factor to $V_{\lambda_i}$ is zero"... – Kuo Jan 27 '24 at 17:52
  • @Kuo That kernel of the restriction of $(A-\lambda_j)^k$ to the generalized eigenspace $V_{\lambda_i}$ for $\lambda_i$ is just ${,v\in V_{\lambda_i}\mid (A-\lambda_j)^k(v)=0,}$, in other words the intersection of that generalized eigenspace with the kernel of $(A-\lambda_j)^k$. And it has dimension$~0$. – Marc van Leeuwen Jan 27 '24 at 18:04
  • I see. Or reading like this $(A-\lambda_i)^{m_i}(A-\lambda_j I)^n v_i = 0$ for that sentence. Math expression is clearer. Thanks for quick reply. – Kuo Jan 27 '24 at 18:23