I'm working with a symmetric matrix and just want to convince myself that, in this specific case, I can say that the trace of this matrix is equal to the sum of its eigenvalues.
Wikipedia says that the trace of any set of matrix products is invariant under cyclic permutation, but the goes on a few lines down to say
Arbitrary permutations are not allowed: in general,
$$Tr(ABC) \not = Tr(ACB)$$
This is a little confusing since this exact example is actually a cyclic permutation ($C$ and $B$ are cycled while $A$ is held fixed).
I just want to clarify whether or not this is a small snafu, or I am totally misunderstanding the details here. I want to use this theorem to say that $Tr(A) = Tr(PDP^{-1}) = Tr(D) =\sum \lambda_i$, and then march happily on my way to try applying this to Principle Component Analysis.
$\text{Trace}(P(DP^{-1})) = \text{Trace}((DP^{-1}) P) = \text{Trace}(D)$. Yes, the trace is the sum of the eigenvalues.