# Aufgaben:Problem 5

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

## Contents

Determine the center of the algebra of complex d × d-matrices. Hint: the center of an algebra $$A$$ is defined as $$Z(A) = \{\phi \in A | \phi \psi = \psi \phi \; \forall \psi \in A \}$$.

## 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 $$M \in \{\lambda\mathbb{I}\}$$.

## Problem 5 (Musterloesung Illmanen)

Let $$E_{ij} \in \mathbb{K}^{n\times n}$$ be the matrix which has the entry 1 at $$(i,j)$$ and 0 else. From proposition $$AE_{ij} = E_{ij}A ~~ \forall 1 \leq i, j \leq n$$. Let $$1 \leq i, j \leq n$$ be fixed: $$AE_{ij}$$ consists solely out of zeros except for the j-th column which is equal to $$(a_{1i}, \ldots , a_{ni})^T$$. So we get $$a_{ii} = a_{jj} ~~ \wedge ~~ a_{ki} = 0 ~~\forall k \neq j$$

We can do this for all indices and this leads to $$a_{ij} = 0 \forall i \neq j ~~\wedge ~~ a_{11} = a_{22} = \ldots = a_{nn}$$. Therefor $$A$$ has to have the form $$\lambda E_n, \lambda \in \mathbb{K}$$

### Ilmanen's solution componentwise

Let $$A \in Z(\mathrm{Mat}_d(\mathbb{C}))$$.

Let $$E_{ij}$$ be the $$d \times d$$-matrix with $$(E_{ij})_{kl} = \delta_{ik} \delta_{jl}$$.

Now let $$1 \leq i, j \leq d$$ with $$i \neq j$$ be fixed and consider $$(E_{ij} A)_{kl} = \sum \limits_{m}(E_{ij})_{km} A_{ml} = \sum \limits_{m}\delta_{ki} \delta_{jm} A_{ml} = \delta_{ki} A_{jl}.$$ Since $$A$$ commutes with all compex $$d \times d$$-matrices, this is the same as $$(A E_{ij})_{kl} = \sum \limits_{m}A_{km} (E_{ij})_{ml} = \sum \limits_{m} A_{km} \delta_{mi} \delta_{jl} = \delta_{jl} A_{ki}.$$

Thus, we have that $$\delta_{ki} A_{jl} = \delta_{jl} A_{ki}$$ Now set $$k=i, l=j$$. We then find $$A_{ii} = A_{jj}$$ Next, set $$k=l=i$$. Still, $$j \neq i$$. It follows that $$A_{ji} = 0.$$ As this holds for an arbitrary choice of $$i \neq j$$, the required form for $$A$$ is $$A = \lambda \mathbb{I}_d$$ for some $$\lambda \in \mathbb{C}$$. And obviously $$\forall \lambda \in \mathbb{C}: \lambda \mathbb{I} \in Z(\mathrm{Mat}_d(\mathbb{C}))$$.