Aufgaben:Problem 5

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An alternative solution can be found here (3.). [2]

Problem 5 (Sheldon)

Let big bold letters denote elements of \(\mathbb{C}^{d \times d}\) and small Greek letters complex numbers.

Let \(M \in Z\)

We make the following observations and conclude that \( Z = \{\lambda\mathbb{I}\}\)

Obviously, the elementes of the center are closed under scalar multiplication (they even form a vector space over \(\mathbb{C}\)). \(\mathbb{I} \in Z \Rightarrow \{\lambda\mathbb{I}\} \subseteq Z\)

The theorem about Jordan decomposition states that \(M\) is conjugate to a matrix in Jordan normal form \(J\) and \(J = CMC^{-1}=MCC^{-1}=M\).

\(M\) is a diagonal matrix.

Suppose to the contrary that \(M\) has a '1' in the superdiagonal, then \(2M\) is supposed to be in the center as well, but it can't, because it is not in Jordan normal form.

All diagonal entries of \(M\) are equal.

Let \( M =\begin{pmatrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & &\ddots& \\ 0 & \ldots & 0 & \lambda_d \end{pmatrix} \) and \( E_{ij} \) be the matrix with zero entries except for a '1' in the i-th row and j-th column.

\(\lambda_i E_{ij} = ME_{ij} = E_{ij}M = \lambda_j E_{ij} \Rightarrow \lambda_i = \lambda_j \)

Because \( 1 \leq i \leq j \leq d \) can be chosen arbitrary, this proves the other inclusion \(M \in \{\lambda\mathbb{I}\}\).