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I'm looking to prove this statement:

Let $\Bbb{F}$ be a field with characteristic $0$. Let $A$ be a $n \times n$ matrix over $\Bbb{F}$. Show that if all the eigenvalues of $T$ are $0$ (in $\Bbb{F}$), then $T$ is nilpotent.

Some sources (such as this and this) use the approach of claiming the characteristic polynomial $\chi(t) \in \Bbb{F}[t]$ of $T$ equals to $t^n$, but the claim is either not justified or their justification requires constraints on $\Bbb{F}$ (specifically, that it should be algebraically closed, so $\chi(t)$ splits and whatnot).

Here is my argument so far:

Since the only eigenvalue of $T$ is $0$, $0$ is the only root of $\chi(t)$. Thus, $\chi(t)$ can be factorized completely over $\Bbb{F}$ as $f_1(t)^{m_1} ... f_k(t)^{m_k} t^{m_0}$ for some $k \in \Bbb{Z}^+_0$, $m_0, m_1, ..., m_k \in \Bbb{Z}^+$ such that $\sum_{i=0}^k m_i = n$, and (monic and irreducible) polynomials $f_1 ... f_k \in \Bbb{F}[t]$ with no roots in $\Bbb{F}$.

I know the minimal polynomial $m(t) \in \Bbb{F}[t]$ of $T$ is equal to $f_1(t)^{l_1} ... f_k(t)^{l_k} t^{l_0}$ for some $l_0, l_1, ..., l_k \in \Bbb{Z}^+$ such that $l_i \leq m_i \forall i$ from $0$ to $k$.

How, then, do I show that $k=0$, or $f_1 = ... = f_k = 1$?

Thank you very much.

Jon
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  • You can consider the matrix $A=Diag(0, S(f))$, where $S(f)$is a companion matrix of irreducible polynomial of degree greater then 1. Then $A$ is not nilpotent. – Mikhail Goltvanitsa Apr 22 '18 at 08:19
  • How do you define the eigenvalues of the matrix? Not as roots of the characteristic polynomial? – Joppy Apr 22 '18 at 11:13

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The statement is wrong: Consider $\mathbb F=\mathbb{R}$ and $$A=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}$$ then we find $\chi_A(t)=t(t^2+1)$, hence $A$ has only $0$ as eigenvalue. However, $A$ is not nilpotent: For $e_1=(1,0,0)$, we find $$A^ke_1=e_1$$ for all $k\in 4\mathbb N$.

asdq
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  • Thank you very much! I've received feedback on this question since asking it here, and the correct constraint on F should be that it's algebraically closed. – Jon Apr 22 '18 at 16:11