Let $x\in \operatorname{gl}(n,F)$ have $n$ distinct eigenvalues $a_1,\ldots,a_n$ in $F$. Prove that the eigenvalues of $\text{ad }x$ are precisely the $n^2$ scalars $a_i-a_j$ ($1\leq i,j\leq n$), which of course need not be distinct.

So we can represent $x$ by an $n\times n$ matrix $X$. We have $Xv_1=a_1v_1,\ldots, Xv_n=a_nv_n$ for eigenvectors $v_1,\ldots,v_n$.

Now, $\operatorname{ad}x$ takes $y\in \operatorname{gl}(n,F)$ to $xy-yx$. I need to show that some $y$ is taken to a scalar multiple of itself, where that scalar is $a_i-a_j$. What could be that $y$?

  • $\begingroup$ Is $X$ assumed to be symmetric by any chance? $\endgroup$ Aug 10, 2013 at 17:14
  • 1
    $\begingroup$ When you type \operatoranme{ad} instead of \text{ad } (with a space between "ad" and the right brace) then it not only prevents italicization but also provides proper spacing in expressions like $a\operatorname{ad}b$. (I think maybe in normal LaTeX as opposed to the stripped-down thing used on this site, it may also prevent line breaks and page breaks. But I'm not sure of that.) $\endgroup$ Aug 10, 2013 at 17:15
  • $\begingroup$ @BranimirĆaćić I don't see that assumption anywhere in the problem statement. $\endgroup$
    – PJ Miller
    Aug 10, 2013 at 17:15
  • $\begingroup$ @MichaelHardy I see, that's good to know $\endgroup$
    – PJ Miller
    Aug 10, 2013 at 17:16
  • $\begingroup$ related: math.stackexchange.com/q/150272/173147 $\endgroup$
    – glS
    Jun 13, 2021 at 16:18

2 Answers 2


Note that the ($n \times n$)-matrix $x$ is diagonalizable since the it has $n$ pairwise different eigenvalues.

Suppose first that $x$ is already a diagonal matrix, say \begin{equation} \tag{1} \label{diagonal form} x = \begin{pmatrix} a_1 & & \\ & \ddots & \\ & & a_n \end{pmatrix}. \end{equation} For the standard basis $\{ e_{ij} \}_{i,j = 1, \dotsc, n}$ of $\mathfrak{gl}(n,F)$ we then have that $$ x e_{ij} = a_i e_{ij} \quad\text{and}\quad e_{ij} x = a_j e_{ij} \qquad \text{for all $i,j$}, $$ and therefore that $$ [x, e_{ij}] = x e_{ij} - e_{ij} x = (a_i - a_j) e_{ij} \qquad \text{for all $i,j$}. $$ This shows that $\operatorname{ad} x$ is diagonalizable with eigenvalues $a_i - a_j$.

In the more general case that $x$ is (only) diagonalizable, there exists $s \in \operatorname{GL}(n,F)$ such that $x = s y s^{-1}$, where $y$ denotes the diagonal matrix from \eqref{diagonal form}. Note that the conjugation map $$ c \colon \mathfrak{gl}(n,F) \to \mathfrak{gl}(n,F), \quad z \mapsto s z s^{-1} $$ is a Lie algebra automorphism with $x = c(y)$. The claimed statement now follows from the previous special case in at least two ways:

  • We can move around the previous eigenvectors by conjugation: The elements $e'_{ij} \in \mathfrak{gl}(n,F)$ given by $$ e'_{ij} := c(e_{ij}) \qquad \text{for all $i,j$} $$ form a basis of $\mathfrak{gl}(n,F)$, for which it follows from $[y, e_{ij}] = (a_i - a_j) e_{ij}$ that $$ [x, e'_{ij}] = [c(y), c(e_{ij})] = c([y, e_{ij}]) = (a_i - a_j) c(e_{ij}) = (a_i - a_j) e'_{ij}. $$

  • The endomorphisms $\operatorname{ad} x$ and $\operatorname{ad} y$ are similar, since it follows from $$ (\operatorname{ad} y) \circ c = (\operatorname{ad} c(x)) \circ c = [c(x), c(-)] = c([x, -]) = c \circ (\operatorname{ad} x) $$ that $\operatorname{ad} x = c^{-1} \circ (\operatorname{ad} y) \circ c$. Since $\operatorname{ad} y$ is diagonalizable with eigenvalues $a_i - a_j$ the same follows for $\operatorname{ad} x$.

  • $\begingroup$ I know this post is old. If someone see this, can explain me what means the notation "$c(-)$"? It's correct and it's formal? $\endgroup$
    – Tryncha
    Mar 10, 2023 at 22:53
  • $\begingroup$ @Tryncha $c$ is a function, so $c(x)$ is just the value of $c$ at $x$. $\endgroup$ Mar 11, 2023 at 13:52
  • $\begingroup$ Yes, but I mean the notation "$c(-)$". It means something like, the value of $c$ at "some arbitrary term not specified"? $\endgroup$
    – Tryncha
    Mar 11, 2023 at 18:17
  • $\begingroup$ +1. Would you please take a look at my generalization of this question math.stackexchange.com/q/4887677/64809 ? $\endgroup$
    – Hans
    Mar 26 at 5:24

First, since $X$ has $n$ distinct eigenvalues, it is diagonalisable, so let $\{e_1,\dotsc,e_n\}$ be a basis for $F^N$ consisting of eigenvectors for $X$, with $X e_k = a_k e_k$.

Next, since $X^T$ has the same eigenvalues as $X$ with the same multiplicities, it is diagonalisable, so let $\{f_1,\dotsc,f_n\}$ be a basis for $F^N$ consisting of eigenvectors for $X^T$, with $X^T f_k = a_k f_k$.

Now, check that $\{e_i f_j^T\}_{i,j=1}^n$ is a basis for $M_n(F)$. What is $(\operatorname{ad}X) \left(e_i f_j^T\right)$ for each $i$ and $j$?

Note: This construction of a basis for $M_n(F)$ is actually quite natural, and even generalises the construction of the standard basis for $M_n(F)$ from the standard basis of $F^n$. In general, if $V$ and $W$ are finite-dimensional vector spaces, then $L(W,V) \cong V \otimes W^\ast$ (naturally!), so that if $\{v_j\}$ is a basis for $V$ and $\{\omega_k\}$ is a basis for $W^\ast$ (e.g., the dual basis to a basis $\{w_k\}$ of $W$), then $\{v_j \otimes \omega_k\}$ is a basis for $V \otimes W^\ast$, and in turn can be identified with a basis for $L(W,V)$, i.e., via identifying $v_j \otimes \omega_k$ with the linear transformation $$ w \mapsto \omega_k(w)v_j. $$ In this case, you have $M_n(F) \cong L(F^n,F^n) \cong F^n \otimes (F^n)^\ast$, with $\{v_j\} = \{e_j\}$ and $\{\omega_k\}$ the dual basis to $\{f_k\}$.

  • $\begingroup$ This is really elegant, Branimir. Thank you! $\endgroup$
    – PJ Miller
    Aug 11, 2013 at 2:54

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