If $\text{ad}(A)$ is nilpotent and $\text{tr}(A) = 0$, then $A$ is a nilpotent matrix.

Question

Let $\mathfrak{gl}\left(\mathbb{R}^n\right)$ be the vector space of all linear transformations $T:\mathbb{R}^n \to \mathbb{R}^n$. If $A$ $\in$ $\mathfrak{gl}\left(\mathbb{R}^n\right)$, we can define the linear map

\begin{align*} \text{ad}(A):\mathfrak{gl}\left(\mathbb{R}^n\right)&\to \mathfrak{gl}\left(\mathbb{R}^n\right)\\ B &\mapsto AB-BA. \\ \end{align*}

I would like to know how to prove the following theorem:

Theorem: Let $A$ $\in$ $\mathfrak{gl}\left(\mathbb{R}^n\right)$, such that $\text{ad}(A)$ is nilpotent and $\text{tr}(A) = 0$, then $A$ is nilpotent.

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I was trying to prove the theorem above by induction in the dimension of $\mathbb{R}^n$, but I did not make much progress.

Does anyone have a nice idea?

Answer 1

A quick and dirty way to prove the theorem is to note that the matrix representation of $\operatorname{ad}(A)$ (i.e. $I\otimes A-A^T\otimes I$) is similar to $C=I\otimes A-A\otimes I$, because $A$ is similar to $A^T$. By extending the field to $\mathbb C$, you may further assume that $A$ is upper triangular. Now the rest are straightforward.

Alternatively, as $A$ has zero trace, it suffices to show that all eigenvalues of $A$ over $\mathbb C$ are equal to each other. Suppose the contrary that $A$ has two different eigenvalues $p$ and $q$. Let $(p, u)$ and $(q,v^T)$ be respectively a right eigenpair and a left eigenpair of $A$. Then $v^Tu=0$. Let $B=uv^T$ and consider $\left(\operatorname{ad}(A)\right)^k(B)$. You shall arrive at a contradiction because $\left(\operatorname{ad}(A)\right)^k(B)$, by construction of $B$, is nonzero but $\left(\operatorname{ad}(A)\right)^k$ must eventually become zero because $\operatorname{ad}(A)$ is nilpotent.

Answer 2

There is a general lemma that says that if $\mathfrak{g}$ is a semisimple Lie subalgebra of some $\mathfrak{gl}_n(K)$ (where $K$ is any field of characteristic $0$), then an element $X \in \mathfrak{g}$ is semisimple / nilpotent as element of $End(K^n)$ if and only if $ad_\mathfrak{g}(X)$ is nilpotent/semisimple. See Bourbaki, Groupes et algèbres de Lie, book I, §6 no. 3 prop. 3 and its corollary.

In your case, the condition $tr(A)=0$ means that you can take $\mathfrak{g} = \mathfrak{sl}_n(\Bbb R) \subset \mathfrak{gl}(\Bbb R^n)$.