What are the eigenvalues and eigenvectors of $\operatorname{ad}x$ for non-diagonalizable $x$?

Question

We know the following proposition is true. The proof together with the specification of the eigenvectors of $\operatorname{ad}x$ is here.

Let $x\in \operatorname{gl}(n,F)$ be diagonalizable with $n$ eigenvalues $a_1,\ldots,a_n$ in $F$. The eigenvalues of $\text{ad }x$, where $\operatorname{ad}x(y):=[x,y]=xy-yx$ are precisely the $n^2$ scalars $a_i-a_j$ ($1\leq i,j\leq n$).

What is the result if $x$ is not diagonalizable? We know $x$ can always be transformed into the Jordan canonical form. I solved the cases of $2\times2$ and $3\times3$ Jordan canonical forms. I would like to know the general solution.

Answer 1

The same conclusion holds.

Use the same argument as in the linked post, if $xv=\lambda v$ and $x^t w = \mu w$ where $v$, $w$ are nonzero (eigenvectors), then $vw^t$ is an eigenvector of $\operatorname{ad}(x)$. Hence $\lambda -\mu$ is an eigenvalue of $\operatorname{ad}(x)$.

Now we only need to show if $x$ is nilpotent, then $\operatorname{ad}(x)$ only has eigenvalue $0$, in other words, $\operatorname{ad}(x)$ is nilpotent (this simple fact is used in the proof of the Engel's theorem in Lie algebra):

$\operatorname{L_x}(y):=xy$ and $\operatorname{R_x}(y):=yx$ are both nilpotent and commute, therefore their difference $L_x-R_x=\operatorname{ad}(x)$ is also nilpotent.

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To be slightly more rigorous, let $v_1, \dotsc, v_l$ (resp. $w_1, \dotsc, w_l$) be a linearly independent set of generalized eigenvectors of $x$ (resp. $x^t$), then we claim $v_iw_j^t$ is a linearly independent set of generalized eigenvectors of $\operatorname{ad}(x)$. The linear independence follows from the independence of $v_i$'s and $w_j$'s.

And if $(x-\lambda_i I)^mv_i=0$, $(x^t-\lambda_j I)^n w_j=0$, then we have $$(L_x-R_x - (\lambda_i-\lambda_j)I)^{m+n} (v_iw_j^t) = ((L_x-\lambda_i) - (R_x-\lambda_j))^{m+n}(v_iw_j^t).$$

Using the binomial theorem, each term in the expansion of $((L_x-\lambda_i I) - (R_x-\lambda_j I))^{m+n}$ either contains $(L_x-\lambda_i I)^p$ for $p\ge m$ or $(R_x-\lambda_j I)^q$ for $q\ge n$ as a factor, and in the first case $$(L_x-\lambda_i I)^pv_iw_j^t=O$$ while similarly in the latter $$(R_x-\lambda_j I)^qv_iw_j^t=v_i((x-\lambda_j)w_j)^t=O.$$

Therefore $v_iw_j^t$ is a generalized eigenvector of $\operatorname{ad}(x)$ (corresponding to eigenvalue $\lambda_i-\lambda_j$).

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To answer the question from the comment. Without loss of generality, let's assume $\lambda_i=\lambda_j=0$.

If $x^m v = 0$ and $x^iv\ne0$ for $i=1, 2, \dotsc, m-1$, also $w^tx^n=0$ and $w^tx^i\ne0$ for $i=1, 2, \dotsc, n-1$. Then in the expansion, the only possible nonzero term is $$(-1)^{n-1}{ {m+n-2}\choose {m-1} } L_x^{m-1}R_x^{n-1}vw^t = (-1)^{n-1}{ {m+n-2}\choose {m-1} }(x^{m-1}v)(w^tx^{n-1}).$$

If this is zero, since $(x^{m-1}v), (w^tx^{n-1})$ are nonzero, their product is also not zero. Therefore the scalar ${ {m+n-2}\choose {m-1} }=0$. Hence the base field has positive characteristic. Similarly, if the base field has characteristic $0$, $p=m+n-1$ must be the minimial integer such that $\operatorname{ad}(x)^p vw^t=0$.

Answer 2

@Justauser has given a great answer to this that's the best way to think about it from a hands-on perspective, but there's also a machinery-heavy perspective from which this fact isn't surprising: $\operatorname{ad}$ is an algebraic representation, and algebraic representations carry semisimple (respectively, nilpotent) elements to semisimple (respectively, nilpotent) elements. Indeed, this is why the notion of the Jordan decomposition of an algebraic Lie algebra can be defined intrinsically (as opposed to an abstract Lie algebra, where we'd want to consider every element of $\operatorname{Lie}(\mathbb R)$ nilpotent and every element of $\operatorname{Lie}(\mathbb R^\times)$ semisimple, but the non-algebraic isomorphism $\operatorname{Lie}(\exp) : \operatorname{Lie}(\mathbb R) \to \operatorname{Lie}(\mathbb R^\times)$ forces us to give up on morphisms respecting this notion). The relevant result in full generality is Theorem 4.4 of Borel - Linear algebraic groups; and the special case in which you are interested is Proposition 1.24 of version 2.00 of Milne - Lie algebras, algebraic groups, and Lie groups.