What are the eigenvalues of $\operatorname{ad}x$?

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 $\operatorname{ad} x$.

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 $\{f_k\}$.