Consider the Lie Algebra $A_1$ having basis {e,h,f} satisfying $[e,f]=h, [h,f]=-2f, [h,e]=2e$. I want to find eigenvalues and eigenvectors of $ad_{(e-f)}$, where $ad$ is the adjoint map. As find as I understand, then I want to those $x\in A_1$, $\lambda\in\mathbb{C}$ st. $\lambda x=ad_{(e-f)}x=[e-f,x]=[e,x]-[f,x]$. But as $[e,x]\in \{h,-2e\}$ and $[f,x]\in\{ 2f,-h\}$ and the only commutator relation giving $h$ is $[e,f]$, then I don't see what would solve this.
Am I misunderstanding something? Are there any specific approach to finding eigenvalues and eigenvectors in such a problem?
Hint: This is basically a linear algebra exercise. With respect to the basis $(e,h,f)$, the linear map $ad_e$ has matrix $\pmatrix{0&-2&0\\0&0&1\\0&0&0}$ and the linear map $ad_f$ has matrix $\pmatrix{0&0&0\\-1&0&0\\0&2&0}$, consequently $ad_{e-f}$ has matrix $\pmatrix{0&-2&0\\1&0&1\\0&-2&0}$. You'll certainly find the characteristic polynomial and hence the eigenvalues of this matrix.
If you did it right, one of those three eigenvalues is $\lambda =2i.$ Now to find a corresponding eigenvector, write down a general element $a_1 e + a_2 h + a_3 f$ in your Lie algebra, and we need to find $a_1,a_2,a_3 \in \mathbb C$ which make
$$[e-f, a_1 e + a_2 h + a_3 f] \stackrel{!}= 2ia_1 e + 2ia_2 h + 2ia_3 f$$
true. Compute the left hand side with the given relations, compare coefficients, get three equations in three variables, solve. Naturally, your solution will not be unique, but all scalar multiples of it will also be.
One eigenvector to $\lambda=2i$ is $25i e +25h+25if$.