# If $E$ has eigenvalues $\lambda_1, \ldots, \lambda_n$, then $B \mapsto [E,B] = EB - BE$ has eigenvalues $\lambda_i - \lambda_j$.

Let $E \in M_n(C)$ be an $n \times n$ matrix with entries in a(n algebraically closed, characteristic 0) field $C$, with eigenvalues $\lambda_1, \ldots, \lambda_n$. Show that the commutator map $M_n(C) \to M_n(C)$ given by $B \mapsto [E,B] = EB - BE$ has eigenvalues $\lambda_i - \lambda_j$.

This is an ingredient a proof of the fact that any regular singular equation $y' = Ay$ over $C((z))$ is equivalent to $v' = Dz^{-1}v$ with $D$ a constant matrix. See exercise (7) (a) (i) of Galois Theory of Differential Equations, Algebraic Groups and Lie Algebras by Marius van der Put (page 6 of the PDF).

This is homework, so hints rather than full answers would be appreciated.

Things I have found that may or may not be useful:

• $EB - BE = (\lambda_i - \lambda_j)B$ can be written $(E-\lambda_i)B = B(E- \lambda_j)$.

• $EB$ and $BE$ have the same eigenvalues.

## 1 Answer

Hint 1. You can assume that $E$ is diagonal without loss of generality (well almost, see below).

Hint 2. Use the natural basis $E_{ij}$ for $M_n$ where the latter is the matrix with a $1$ in the $(i,j)$-th entry, zero elsewhere. Calculate the image of $E_{ij}$.

Hint 3. This works by a suitable choice of basis when the matrix is diagonalisable. If not, one requires a rather more delicate analysis, using the Jordan normal form.