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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

3 Answers3

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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
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If a matrix is in Jordan normal form, then it is in particular block diagonal. But it is easy to see that the minimal polynomial of any block diagonal matrix is the least common multiple of the minimal polynomials of its block submatrices.

But if $J_k(\lambda)$ denotes a Jordan block of size $k$ with diagonal entries all equal to $\lambda$, it is easy to check that its mimimal polynomial is $(t-\lambda)^k$. It therefore follows that the minimial polynomial is $\prod_{j=1}^m (t-\lambda_j)^{r_j}$ where the $\lambda_j$ $(1 \leq j \leq m)$ are the distinct eigenvalues of the matrix, and $r_j$ is the size of the largest Jordan block associated to the eigenvalue $\lambda_j$.

krm2233
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