Difference between revisions of "Talk:Aufgaben:Problem 5"

Revision as of 14:06, 28 July 2015

Does anyone like tensor notation and wants to tell me whether this is formally correct?

--Nik (talk) 14:41, 28 July 2015 (CEST)

Ilmanen's solution in Einstein notation

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

Let \(E_{ij}\) be the \(d \times d\)-matrix with \((E_{ij})^k{}_l = \delta^{ij} \delta_{kl} \).

Now, consider $$(E_{ij} A)^k{}_l = (E_{ij})^k{}_m A^m{}_l = \delta^k{}_i \delta^j{}_m A^m{}_l = \delta^k{}_i A^j{}_l$$ but since \(A\) commutes with all compex \(d \times d\)-matrices, this is the same as $$(A E_{ij})^k{}_l = A^k{}_m (E_{ij})^m{}_l = A^k{}_m \delta^m{}_i \delta^j{}_l = \delta^j{}_l A^k{}_i$$

Thus, we have that $$\delta^k{}_i A^j{}_l = \delta^j{}_l A^k{}_i$$ As this holds for any \(1 \leq i,j,k,l \leq d\), we find: $$\forall i \neq j: A^i{}_j = 0 \ \text{and} \ A^i{}_i = A^j{}_j$$ which requires that \(A\) takes the form $$A = \lambda \mathbb{I}_d$$ for some \(\lambda \in \mathbb{C}\)

Why do you write this with up down indices at all? Wouldn't it be normal to just use down indices, as these are just matrix multiplications we are dealing with, or am I missing something?

Carl (talk) 16:06, 28 July 2015 (CEST)