eigenvalues of adjoint operator of a diagonizable complex matrix

Let $A \in M_{n}(\mathbb{C})$ with eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$. Define a linear $\operatorname{map} \operatorname{ad}(A): M_{n}(\mathbb{C}) \longrightarrow M_{n}(\mathbb{C})$ by $\operatorname{ad}(A)(B)=A B-B A$ for $B \in M_{n}(\mathbb{C})$. Determine the eigenvalues of $\operatorname{ad}(A)$ in terms of $\lambda_{1}, \ldots, \lambda_{n}$.

Here, we assume that $A$ is diagonizable.

Previously, I worked on the following question:

Let $W$ be the subspace of $M_2(\mathbb R)$ consisting of $2 \times 2$ matrices with trace $0$. Let $V$ be the space of linear maps from $W$ to $W$. Define a linear map $T : W \rightarrow V$ by $A \mapsto \operatorname{ad}(A)$, where $\operatorname{ad}(A)(B) = AB - BA$ for $B \in W$. Choose bases for $W$ and $V$ and compute the matrix of $T$ with respect to the bases.

and solved it like this:

First, we introduce a base for $W$ as follows: \begin{equation*} \mathcal{B}_W= \bigg \{ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \bigg \}. \end{equation*} Now, for $V$ we have the following set as a base with respect to $\mathcal{B}_W$: \begin{equation*} \begin{aligned} \mathcal{B}_V=\{T_{ij}:= &\text{The $3 \times 3$ matrix that the only nonzero element} \\ &\text{of it is the $ij$-th element, that equals $1$} \mid 1 \leq i, j \leq 3 \} \end{aligned} \end{equation*} Now, to compute the matrix of $T$ with respect to these bases, first note that for any $1 \leq i \leq 3$, we have $\alpha_i . \alpha _i=0$. So, without loss of generality, writing the matrix $A$ as $(\alpha_1, \alpha_2, \alpha_3)$, and $B=(a\alpha_1, b \alpha_2, c\alpha_3)$, we have: \begin{equation*} \begin{aligned} T\alpha_1 (a, b, c) &= (2c, 0, -b), \\ T\alpha_2 (a, b, c) &= (0, -2c, a), \\ T\alpha_3 (a, b, c) &= (-2a, 2b, 0). \end{aligned} \end{equation*} Now, writing this in forms of matrices, we'll have: \begin{equation*} \begin{aligned} T\alpha_1&= \begin{bmatrix}0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & -1 & 0 \end{bmatrix},\\ T\alpha_2&= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -2 \\ 1 & 0 & 0 \end{bmatrix},\\ T\alpha_3&= \begin{bmatrix}-2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}.\\ \end{aligned} \end{equation*} So, the matrix of $T$ is going to be a $9 \times 3$ matrix as following: \begin{equation*} \begin{bmatrix} 0 & 0 & -2\\ 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & -2 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}. \end{equation*}

Now, I think to solve the eigenvalues question, I should be able to use the second question I mentioned, however I don't know how and where to start. any help's appreciated.

It looks like you are mixing two things. Your map $T$ takes $A$ to $\operatorname{ad}(A)$. That bears little relation to the eigenvalues of $\operatorname{ad}(A)$, which is what you are asking about.

And this is just a straight computation. The fact that $A$ is diagonalizable allows you to assume that $A$ is diagonal with diagonal $\lambda_1,\ldots,\lambda_n$. Then $$ (AB-BA)_{kj}=(\lambda_k-\lambda_j)B_{kj}. $$ In coordinates, then, the eigenvalue equation $\operatorname{ad}(A)B=\alpha B$ looks $$ (\lambda_k-\lambda_j)B_{kj}=\alpha\,B_{kj}. $$ So as soon as some $B_{kj}\ne0$, this forces $\alpha=\lambda_k-\lambda_j$. Choosing $B$ to be the matrix $E_{kj}$ that has $1$ in the $k,j$ entry and zeroes elsewhere, you get $\operatorname{ad}(A)B=(\lambda_k-\lambda_j) B$. Thus the eigenvalues of $\operatorname{ad}(A)$ are $$\big\{\lambda_k-\lambda_j:\ k,j\in \{1,\ldots,n\}\big\}.$$