# İf x is diagonalizable then ad(x) is also diagonalizable

I start to study lie algebras from K. Erdmann, Mark J. Wildon-Introduction to Lie Algebras and I try to solve question below but actually I can't see. How can I start ? Give me a hint please

Let $$V$$ be an $$n$$-dimensional complex vector space and let $$L = gl(V)$$. Suppose that $$x \in L$$ is diagonalisable, with eigenvalues $$λ_1, . . . , λ_n$$. Show that $$\text{ad}(x) \in gl(L)$$ is also diagonalisable and that its eigenvalues are $$λ_i − λ_j$$ for $$1 ≤ i, j ≤ n$$. Definition: Let $$L$$ be a Lie algebra $$\text{ad}(x)$$ is a linear map from $$L$$ to itself defined by $$\text{ad}(x)(y)=[x,y]$$ :

## 1 Answer

Take $\mathcal B$ a basis wherein $x$ is diagonalizable and denote $A$ its matrix. Let $B$ the matrix of $y$. $\lambda$ is an eigenvalue of $\operatorname{ad}(x)$ if

$$AB-BA=\lambda B$$ so using the components we get

$$\lambda_i b_{ij}-b_{ij}\lambda_j=\lambda b_{ij}\implies \lambda=\lambda_i-\lambda_j \quad\text{if}\; b_{ij}\ne0$$ so we see that the matrix $E_{ij}$ with all the entries are $0$ except the $(i,j)$-entry which is equal $1$ is an eigenvector associated to the eigenvalue $\lambda_i-\lambda_j$ and since $(E_{ij})$ is a basis for $\mathcal M_n(\Bbb R)$ then $\operatorname{ad}(x)$ is diagonalizable.