Difference between revisions of "Talk:Aufgaben:Problem 5"
(Ilmanen's solution a bit more rigorous) |
(added tensor index spacing) |
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Let \(A \in Z(\mathrm{Mat}_d(\mathbb{C}))\). | Let \(A \in Z(\mathrm{Mat}_d(\mathbb{C}))\). | ||
| − | Let \(E_{ij}\) be the \(d \times d\)-matrix with \((E_{ij})^ | + | Let \(E_{ij}\) be the \(d \times d\)-matrix with \((E_{ij})^k{}_l = \delta^{ij} \delta_{kl} \). |
Now, consider | Now, consider | ||
| − | $$(E_{ij} A)^ | + | $$(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 | but since \(A\) commutes with all compex \(d \times d\)-matrices, this is the same as | ||
| − | $$(A E_{ij})^ | + | $$(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$$ |
Thus, we have that | Thus, we have that | ||
| − | $$\delta^ | + | $$\delta^k{}_i A^j{}_l = \delta^j{}_l A^k{}_i$$ |
As this holds for ''any'' \(1 \leq i,j,k,l \leq d\), we find: | As this holds for ''any'' \(1 \leq i,j,k,l \leq d\), we find: | ||
| − | $$\forall i \neq j: A^ | + | $$\forall i \neq j: A^i{}_j = 0 \ \text{and} \ A^i{}_i = A^j{}_j$$ |
which requires that \(A\) takes the form | which requires that \(A\) takes the form | ||
$$A = \lambda \mathbb{I}_d$$ | $$A = \lambda \mathbb{I}_d$$ | ||
for some \(\lambda \in \mathbb{C}\) | for some \(\lambda \in \mathbb{C}\) | ||
Revision as of 14:02, 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 \(\lambda \in \mathbb{C}\)
