What are the eigenvalues of $\operatorname{ad}x$? 
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 $y$?
 A: 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 $\{f_k\}$.
A: 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 $\operatorname{ad} x$.
