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Let $F$ be a field, $A : V \to V$ a nilpotent endomorphism in a finite dimensional vector space. Proof, that the Minimal Polynomial $m_A$ equals the characteristic polynomial $f_A$ if and only if for each factorization $f_A (x) = p(x)q(x)$, $Im(p(A)) = Ker(q(A))$.

I am stuck with both directions. I know, that the minimal polynomial of A must be $x^k$ for a $k$, so each factorization must be $x^i \cdot x^{k-i}$. I would appreciate hints as well as a solution.

Luis
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  • In one direciton, if the minimal properly divides the characteristic, what happens if $p(x)$ is the minimal polynomial? If they are equal, think about the Jordan form. – Arturo Magidin Jul 21 '21 at 17:13
  • P.S. Note that you are factoring the characteristic polynomial, not the minimal one. If $\dim(V)=n$, then $f_A(x)=x^n$, and $m_A(x)=x^k$ with $1\leq k\leq n$. The factorization is actually $x^rx^{n-r}$. – Arturo Magidin Jul 21 '21 at 17:20
  • we factorize $x^n = x^r x^{n-r}$, so $Im(A^r) = {0} = Ker(A^{n-r})$, so rank$(A^{n-r})=n$ which is a contradiction, because $A$ is nilpotent. – Luis Jul 21 '21 at 18:43
  • and for the other direction, we know, that if $m_A (x)= x^n = x^r x^{n-r}$, so $A^r \neq 0 \neq A^{n-r}$ for all r. So if we take $v \in Im(A^r)$, $A^n v = 0$, so $v \in Ker(A^{n-r})$. – Luis Jul 21 '21 at 18:59
  • That only gives you one inclusion. You also need to show $\mathrm{ker}(A^{n-r}) \subseteq \mathrm{Im}(A^r)$. – Arturo Magidin Jul 21 '21 at 19:02
  • By the way: it is better to put your efforts into the post, not the comments. – Arturo Magidin Jul 21 '21 at 19:03

1 Answers1

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As you note, because $A$ is nilpotent, the characteristic polynomial is $f_A(x)=x^n$ with $n=\dim(V)$ (assuming your characteristic polynomials are monic; otherwise, it is $(-1)^nx^n$). THe minimal polynomial is $m_A(x) = x^k$, with $1\leq k\leq n$.

Express $f_A(x) = m_A(x)x^{n-k}$. Then $\mathrm{Im}(m_A(A))=\mathrm{ker}(A^{n-k})$. But $m_A(A)$ is the zero linear transformation, so the image is trivial. Thus, $\mathrm{ker}(A^{n-k})$ is trivial, so $A^{n-k}$ is one-to-one. This is impossible if $n-k\gt 0$ (since $A$, being nilpotent, has nontrivial kernel), thus $n-k=0$ and hence $m_A(x)=f_A(x)$.

Conversely, suppose that $m_A(x)=f_A(x)$. Let $\beta=\{v_1,ldots,v_n\}$ be a Jordan canonical basis for $A$; that is, $A(v_m) = v_{m-1}$ if $m=2,\ldots,n$, and $A(v_1)=\mathbf{0}$. Then it is easy to verify that $\mathrm{Im}(A^r) = \mathrm{span}(v_1,\ldots,v_{n-r})$, and $\mathrm{ker}(A^s) = \mathrm{span}(v_1,\ldots,v_s)$. Thus, if $f_A(x) = x^rx^{n-r}$, we have $$\mathrm{Im}(A^r) = \mathrm{span}(v_1,\ldots,v_{n-r}) = \mathrm{ker}(A^{n-r}),$$ as required.

Related, this question characterizing the (finite dimensional) linear operators for which the minimal and characteristic polynomials agree.

Arturo Magidin
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