Trace equals sum of eigenvalues ↩

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created: 2021-10-19 12:35:53
modified: 2022-01-10 04:13:04

If are the eigenvalues of matrix , then:

^6917e2

^b99046
## Proof
If is diagonalizable, then there exists an for which:

By definition , and since :

Here we used that the trace is invariant to cyclic permutations on products.