1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.